All practice questions

Question 4281Question →

If $\frac{4 + 3\sqrt{2}}{3 - 2\sqrt{2}}$ is expressed in the simplified form $a + b\sqrt{2}$ , where $a$ and $b$ are integers, what is the value of $a + b$ ?

Show answer & explanation

Answer: 41

Answer

The value of

a + b

is 41.

To rationalise

\frac{4 + 3\sqrt{2}}{3 - 2\sqrt{2}}

, multiply both numerator and denominator by the conjugate

3 + 2\sqrt{2}

. The denominator becomes

3^2 - (2\sqrt{2})^2 = 9 - 8 = 1

. Expanding the numerator gives

(4)(3) + 4(2\sqrt{2}) + 3\sqrt{2}(3) + 3\sqrt{2}(2\sqrt{2}) = 12 + 8\sqrt{2} + 9\sqrt{2} + 12 = 24 + 17\sqrt{2}

. Comparing with

a + b\sqrt{2}

gives

a = 24

and

b = 17

, so

a + b = 24 + 17 = 41

.

Step-by-Step Solution

1

Multiply the numerator and denominator by the conjugate of the denominator.

The expression becomes

\frac{(4 + 3\sqrt{2})(3 + 2\sqrt{2})}{(3 - 2\sqrt{2})(3 + 2\sqrt{2})}

.

Multiplying by the conjugate eliminates surds from the denominator using the difference of two squares identity

(x-y)(x+y) = x^2 - y^2

.

2

Simplify the denominator.

(3)^2 - (2\sqrt{2})^2 = 9 - (4 \times 2) = 9 - 8 = 1

.

Squaring

2\sqrt{2}

yields

2^2 \times (\sqrt{2})^2 = 4 \times 2 = 8

.

3

Expand the numerator.

(4 \times 3) + (4 \times 2\sqrt{2}) + (3\sqrt{2} \times 3) + (3\sqrt{2} \times 2\sqrt{2}) = 12 + 8\sqrt{2} + 9\sqrt{2} + 12 = 24 + 17\sqrt{2}

.

Applying the distributive law and grouping rational terms together and like surd terms together.

4

Identify the values of

a

and

b

and calculate

a + b

.

a = 24

,

b = 17

, so

a + b = 24 + 17 = 41

.

Matching coefficients of the simplified surd form

a + b\sqrt{2}

.

Key Concept

Rationalisation of binomial surd denominators using conjugates

Estimated Time:2m 0s

Question 4282Question →

The total profit $P(x)$ , in thousands of Naira, obtained from producing and selling $x$ hundred units of a commodity is modeled by the function $P(x) = -x^3 + 6x^2 + 15x - 8$ , where $x \ge 0$ . What is the maximum profit achievable?

Show answer & explanation

Answer: 92

Answer

The maximum profit achievable is 92 thousand Naira.

To find the maximum profit, we find the stationary points of

P(x) = -x^3 + 6x^2 + 15x - 8

by taking the derivative

P'(x) = -3x^2 + 12x + 15

and setting it to 0. Solving

-3(x-5)(x+1) = 0

with

x \ge 0

yields

x = 5

. Testing the second derivative gives

P''(5) = -18 < 0

, confirming

x = 5

is a maximum. Substituting

x = 5

into

P(x)

yields

P(5) = 92

.

Step-by-Step Solution

1

Differentiate the profit function

P(x)

with respect to

x

P'(x) = -3x^2 + 12x + 15

Stationary points occur where the rate of change of profit (the derivative) is equal to zero.

2

Set

P'(x) = 0

and solve for

x

-3(x^2 - 4x - 5) = 0 \implies (x - 5)(x + 1) = 0

, giving critical values

x = 5

and

x = -1

Factoring the quadratic equation yields the critical values of production level.

3

Filter critical values based on physical domain constraints

x = 5

(reject

x = -1

since production

x \ge 0

)

Production quantities cannot be negative in physical real-life contexts.

4

Perform the second derivative test to confirm the nature of the stationary point

P''(x) = -6x + 12 \implies P''(5) = -6(5) + 12 = -18

Since

P''(5) < 0

, the function achieves a local maximum at

x = 5

.

5

Substitute

x = 5

back into original function

P(x)

P(5) = -(5)^3 + 6(5)^2 + 15(5) - 8 = -125 + 150 + 75 - 8 = 92

Evaluating

P(5)

gives the maximum total profit value.

Key Concept

Optimization and Maximum Values using First and Second Derivatives

Question 4283Question →

If the line $3x + py - 7 = 0$ is perpendicular to the line passing through the points $(1, -2)$ and $(4, 7)$ , what is the value of $p$ ?

9

-1

-9

3

Show answer & explanation

Answer: $9$

Answer

The value of

p

is

9

.

The line passing through

(1, -2)

and

(4, 7)

has a gradient of

m_1 = \frac{7 - (-2)}{4 - 1} = 3

. The equation

3x + py - 7 = 0

can be rewritten as

y = -\frac{3}{p}x + \frac{7}{p}

, giving a gradient of

m_2 = -\frac{3}{p}

. For perpendicular lines, the product of their gradients must equal

-1

, so

3 \times \left(-\frac{3}{p}\right) = -1

, which simplifies to

p = 9

.

Step-by-Step Solution

1

Calculate the gradient (

m_1

) of the line passing through

(1, -2)

and

(4, 7)

m_1 = \frac{7 - (-2)}{4 - 1} = \frac{9}{3} = 3

The gradient between two points

(x_1, y_1)

and

(x_2, y_2)

is given by

\frac{y_2 - y_1}{x_2 - x_1}

.

2

Express the line

3x + py - 7 = 0

in slope-intercept form (

y = mx + c

) to find its gradient (

m_2

)

py = -3x + 7 \implies y = -\frac{3}{p}x + \frac{7}{p}

, so

m_2 = -\frac{3}{p}

The coefficient of

x

when solved for

y

represents the gradient of the straight line.

3

Apply the perpendicularity condition

m_1 \cdot m_2 = -1

and solve for

p

3 \cdot \left(-\frac{3}{p}\right) = -1 \implies -\frac{9}{p} = -1 \implies p = 9

Two non-vertical lines are perpendicular if and only if the product of their gradients is

-1

.

Key Concept

Perpendicular Lines and Gradients

Estimated Time:1m 30s

Question 4284Question →

What is the value of the definite integral $\int_{1}^{3} (2x + 1) \, dx$ ?

10

12

14

2

Show answer & explanation

Answer: 10

Answer

10

The antiderivative of

2x + 1

is

x^2 + x

. Evaluating at the upper boundary

x = 3

gives

3^2 + 3 = 12

, and at the lower boundary

x = 1

gives

1^2 + 1 = 2

. Subtracting the lower bound result from the upper bound result yields

12 - 2 = 10

.

Step-by-Step Solution

1

Find the indefinite integral (antiderivative) of

2x + 1

\int (2x + 1) \, dx = x^2 + x

Apply the power rule of integration:

\int x^n dx = \frac{x^{n+1}}{n+1}

.

2

Evaluate the antiderivative at the upper limit

x = 3

3^2 + 3 = 9 + 3 = 12

Substitute

x = 3

into

x^2 + x

.

3

Evaluate the antiderivative at the lower limit

x = 1

1^2 + 1 = 1 + 1 = 2

Substitute

x = 1

into

x^2 + x

.

4

Subtract the lower limit value from the upper limit value

12 - 2 = 10

According to the Fundamental Theorem of Calculus:

\int_{a}^{b} f(x)dx = F(b) - F(a)

.

Key Concept

Definite Integrals of Linear Polynomials

Question 4285Question →

An object of mass $0.20\text{ kg}$ executes simple harmonic motion with an amplitude of $0.05\text{ m}$ and a maximum acceleration of $20\text{ m/s}^2$ . What is the speed of the object when its displacement from the equilibrium position is $0.03\text{ m}$ ?

0.16\text{ m/s}

0.60\text{ m/s}

0.80\text{ m/s}

1.00\text{ m/s}

Show answer & explanation

Answer: $0.80\text{ m/s}$

Answer

The speed of the object at a displacement of

0.03\text{ m}

is

0.80\text{ m/s}

.

The maximum acceleration in simple harmonic motion is given by

a_{\text{max}} = \omega^2 A

. Substituting

a_{\text{max}} = 20\text{ m/s}^2

and

A = 0.05\text{ m}

gives

\omega^2 = 400\text{ rad}^2/\text{s}^2

, so

\omega = 20\text{ rad/s}

. The speed at any displacement

x

is given by

v = \omega \sqrt{A^2 - x^2}

. For

x = 0.03\text{ m}

,

v = 20 \sqrt{0.05^2 - 0.03^2} = 20 \times 0.04 = 0.80\text{ m/s}

.

Step-by-Step Solution

1

Determine the angular frequency (

\omega

) of the simple harmonic motion from the maximum acceleration formula.

\omega = 20\text{ rad/s}

Maximum acceleration is given by

a_{\text{max}} = \omega^2 A

. Rearranging gives

\omega^2 = \frac{a_{\text{max}}}{A} = \frac{20}{0.05} = 400\text{ rad}^2/\text{s}^2

, so

\omega = 20\text{ rad/s}

.

2

Calculate the speed (

v

) at the given displacement (

x = 0.03\text{ m}

) using the SHM velocity formula.

v = 0.80\text{ m/s}

The speed at displacement

x

is

v = \omega \sqrt{A^2 - x^2} = 20 \times \sqrt{0.05^2 - 0.03^2} = 20 \times \sqrt{0.0016} = 20 \times 0.04 = 0.80\text{ m/s}

.

Key Concept

Simple Harmonic Motion Velocity and Acceleration Relationships

Estimated Time:1m 30s

Question 4286Question →

A body $A$ of mass $4.0\text{ kg}$ moving due east at a velocity of $6.0\text{ m s}^{-1}$ collides head-on with a body $B$ of mass $2.0\text{ kg}$ moving due west at $3.0\text{ m s}^{-1}$ . If body $A$ continues to move due east after the collision with a speed of $1.0\text{ m s}^{-1}$ , what is the velocity of body $B$ after the collision?

7.0\text{ m s}^{-1}

due East

13.0\text{ m s}^{-1}

due East

7.0\text{ m s}^{-1}

due West

11.0\text{ m s}^{-1}

due East

Show answer & explanation

Answer: $7.0\text{ m s}^{-1}$ due East

Answer

7.0\text{ m s}^{-1}

due East

According to the principle of conservation of linear momentum, the total momentum before collision equals the total momentum after collision. Assigning positive to East and negative to West, total initial momentum is

4(6) + 2(-3) = 18\text{ kg m s}^{-1}

. Equating this to final momentum

4(1) + 2v_B

gives

2v_B = 14

, resulting in

v_B = +7.0\text{ m s}^{-1}

, which represents

7.0\text{ m s}^{-1}

due East.

Step-by-Step Solution

1

Define a reference direction for 1D momentum vectors

Let East be positive (+) and West be negative (-).

Linear momentum is a vector quantity, so opposite directions must have opposite algebraic signs.

2

Calculate the total initial momentum before collision (

p_i

)

p_i = m_A u_A + m_B u_B = (4.0\text{ kg})(+6.0\text{ m s}^{-1}) + (2.0\text{ kg})(-3.0\text{ m s}^{-1}) = 24.0 - 6.0 = 18.0\text{ kg m s}^{-1}

.

Body B moves west, so its initial velocity is

-3.0\text{ m s}^{-1}

.

3

Formulate the total final momentum after collision (

p_f

)

p_f = m_A v_A + m_B v_B = (4.0\text{ kg})(+1.0\text{ m s}^{-1}) + (2.0\text{ kg})v_B = 4.0 + 2.0 v_B

.

Body A moves east after collision, so its final velocity is

+1.0\text{ m s}^{-1}

.

4

Apply the Law of Conservation of Linear Momentum (

p_i = p_f

)

18.0 = 4.0 + 2.0 v_B \implies 2.0 v_B = 14.0 \implies v_B = +7.0\text{ m s}^{-1}

.

Since total initial momentum equals total final momentum in an isolated system.

5

Interpret the sign of the calculated velocity

Since

v_B

is positive (

+7.0\text{ m s}^{-1}

), body B moves at

7.0\text{ m s}^{-1}

due East.

Positive values correspond to the defined East direction.

Key Concept

Principle of Conservation of Linear Momentum in 1D head-on collisions

Question 4287Question →

The electric current passing through a resistor is measured as $(5.0 \pm 0.1)\text{ A}$ , and the resistance is measured as $(10.0 \pm 0.3)\text{ }\Omega$ . What is the maximum percentage error in the calculated electrical power dissipated by the resistor?

7.0\%

5.0\%

4.0\%

12.0\%

Show answer & explanation

Answer: $7.0\%$

Answer

The maximum percentage error in the calculated power is

7.0\%

.

Electrical power is calculated using the formula

P = I^2 R

. In error analysis, the maximum fractional error for a quantity

y = a^n b^m

is

\frac{\Delta y}{y} = n\frac{\Delta a}{a} + m\frac{\Delta b}{b}

. Therefore, the maximum percentage error in

P

is equal to

2 \times (\text{percentage error in } I) + (\text{percentage error in } R) = 2(2.0\%) + 3.0\% = 7.0\%

.

Step-by-Step Solution

1

Calculate the percentage error in the measured current

I

.

\frac{\Delta I}{I} \times 100\% = \frac{0.1}{5.0} \times 100\% = 2.0\%

Percentage error is the ratio of absolute uncertainty to the measured value expressed as a percentage.

2

Calculate the percentage error in the measured resistance

R

.

\frac{\Delta R}{R} \times 100\% = \frac{0.3}{10.0} \times 100\% = 3.0\%

Applying the same percentage error formula to the resistance measurement.

3

Apply the error propagation formula for power

P = I^2 R

.

\frac{\Delta P}{P} \times 100\% = 2\left(\frac{\Delta I}{I} \times 100\%\right) + \left(\frac{\Delta R}{R} \times 100\%\right) = 2(2.0\%) + 3.0\% = 7.0\%

When physical quantities are raised to a power and multiplied, their fractional/percentage errors are multiplied by the exponent and summed to get the maximum error.

Key Concept

Propagation of errors in derived quantities involving powers and products

Alternative Method

Calculate absolute values of minimum and maximum possible power, then compute percentage variation relative to nominal power.

Estimated Time:1m 30s

Question 4288Question →

Find the positive integer value of $n$ such that $^{n}P_4 = 42 \times {^{n}P_2}$ .

Show answer & explanation

Answer: 9

Answer

The positive integer value of

n

is 9.

Expanding

^{n}P_4

as

n(n-1)(n-2)(n-3)

and

^{n}P_2

as

n(n-1)

allows dividing out

n(n-1)

, leading to

(n-2)(n-3) = 42

. Expanding and factoring gives

n^2 - 5n - 36 = 0

, which yields

n = 9

as the only valid positive integer.

Step-by-Step Solution

1

Apply the permutation formula

^{n}P_r = \frac{n!}{(n-r)!}

^{n}P_4 = n(n-1)(n-2)(n-3)

and

^{n}P_2 = n(n-1)

By definition of permutations, selecting

r

items from

n

distinct items without replacement.

2

Substitute the expansions into the given relation

n(n-1)(n-2)(n-3) = 42 n(n-1)

Direct substitution into

^{n}P_4 = 42 \times {^{n}P_2}

.

3

Simplify by dividing out common non-zero terms

(n-2)(n-3) = 42

Since

n \ge 4

,

n(n-1) \neq 0

and can be safely divided from both sides.

4

Form and solve the quadratic equation

n^2 - 5n + 6 = 42 \implies n^2 - 5n - 36 = 0 \implies (n-9)(n+4) = 0

Expanding terms and factoring the resulting quadratic expression.

5

Determine the valid root

n = 9

Permutation total items

n

must satisfy

n \ge r \ge 0

, rejecting the negative root

n = -4

.

Key Concept

Algebraic equations involving permutations

Estimated Time:1m 15s

Question 4289Question →

An electromagnetic wave has a wavelength of $1.5 \times 10^{-7}\text{ m}$ in a vacuum. It enters an optical medium where its speed decreases to $2.0 \times 10^8\text{ m/s}$ . Given that the speed of light in vacuum is $c = 3.0 \times 10^8\text{ m/s}$ , what is the wavelength of the wave in the medium, and to which spectral region does its frequency belong?

1.0 \times 10^{-7}\text{ m}

and Ultraviolet region

1.0 \times 10^{-7}\text{ m}

and Visible light region

2.25 \times 10^{-7}\text{ m}

and Ultraviolet region

1.5 \times 10^{-7}\text{ m}

and Infrared region

Show answer & explanation

Answer: $1.0 \times 10^{-7}\text{ m}$ and Ultraviolet region

Answer

The wavelength in the medium is

1.0 \times 10^{-7}\text{ m}

and its frequency belongs to the Ultraviolet region.

When an electromagnetic wave transitions into a medium, its frequency

f

remains constant at

2.0 \times 10^{15}\text{ Hz}

, placing it strictly in the Ultraviolet spectrum. Its wavelength in the medium shrinks to

\lambda_{med} = \frac{v}{f} = \frac{2.0 \times 10^8}{2.0 \times 10^{15}} = 1.0 \times 10^{-7}\text{ m}

.

Step-by-Step Solution

1

Calculate the frequency of the wave in vacuum using the wave equation

c = f \lambda_{vac}

.

f = \frac{3.0 \times 10^8\text{ m/s}}{1.5 \times 10^{-7}\text{ m}} = 2.0 \times 10^{15}\text{ Hz}

.

Frequency is an intrinsic property determined by the source and does not change when entering a new medium.

2

Identify the electromagnetic spectrum region corresponding to the calculated frequency.

A frequency of

2.0 \times 10^{15}\text{ Hz}

falls in the range

7.5 \times 10^{14}\text{ Hz}

to

3.0 \times 10^{16}\text{ Hz}

, which corresponds to the Ultraviolet region.

Spectral classifications are uniquely determined by frequency (or vacuum wavelength).

3

Determine the wavelength in the medium using

\lambda_{med} = \frac{v}{f}

.

\lambda_{med} = \frac{2.0 \times 10^8\text{ m/s}}{2.0 \times 10^{15}\text{ Hz}} = 1.0 \times 10^{-7}\text{ m}

.

The wavelength changes proportionally with phase velocity in a medium.

Key Concept

Invariance of electromagnetic wave frequency across media boundaries and wave equation relationship

v = f \lambda

.

Estimated Time:2m 0s

Question 4290Question →

A stone is projected from ground level with an initial velocity of $40\text{ m/s}$ at an angle of $30^\circ$ to the horizontal. What is the total time of flight of the stone, in seconds? (Take $g = 10\text{ m/s}^2$ )

Show answer & explanation

Answer: 4

Answer

The total time of flight of the stone is

4\text{ s}

.

The total time of flight

T

for a projectile launched over level ground is calculated using

T = \frac{2 u \sin \theta}{g}

. Substituting

u = 40\text{ m/s}

,

\theta = 30^\circ

, and

g = 10\text{ m/s}^2

yields

T = \frac{2 \times 40 \times 0.5}{10} = 4\text{ s}

.

Step-by-Step Solution

1

Find the vertical component of the launch velocity

u_y = 20\text{ m/s}

The vertical motion determines the time the projectile remains in the air.

2

Calculate the total time of flight

T = 4\text{ s}

Applying

T = \frac{2 u \sin \theta}{g} = \frac{2 \times 20}{10} = 4\text{ s}

gives the total duration before landing back at ground level.

Key Concept

Time of Flight in Projectile Motion

Estimated Time:45s

Question 4291Question →

If $(x, y)$ satisfies the simultaneous equations $x + 2y = 5$ and $x^2 + y^2 = 10$ , what is the positive value of $x$ ?

Show answer & explanation

Answer: 3

Answer

The positive value of

x

is 3.

Isolating

x

in the linear equation gives

x = 5 - 2y

. Substituting this expression into

x^2 + y^2 = 10

yields

(5 - 2y)^2 + y^2 = 10

. Expanding gives

25 - 20y + 4y^2 + y^2 = 10 \implies 5y^2 - 20y + 15 = 0

. Dividing all terms by

5

produces

y^2 - 4y + 3 = 0

, which factors as

(y - 1)(y - 3) = 0

, so

y = 1

or

y = 3

. Substituting these into

x = 5 - 2y

gives

x = 3

when

y = 1

and

x = -1

when

y = 3

. The positive value of

x

is 3.

Step-by-Step Solution

1

Express

x

from the linear equation

x = 5 - 2y

Isolating

x

allows substitution into the quadratic equation.

2

Substitute into the quadratic equation

(5 - 2y)^2 + y^2 = 10

Eliminates variable

x

to create a single-variable equation in

y

.

3

Expand and simplify

5y^2 - 20y + 15 = 0 \implies y^2 - 4y + 3 = 0

Transforms the equation into standard quadratic form for easy factorization.

4

Solve for

y

y = 1 \text{ or } y = 3

Factoring

(y - 1)(y - 3) = 0

yields the two possible values for

y

.

5

Determine corresponding

x

values and select the positive one

x = 3

(from

y = 1

)

Evaluating

x = 5 - 2y

gives

x = 3

and

x = -1

; the positive result requested is 3.

Key Concept

Solving simultaneous linear and quadratic equations by substitution

Estimated Time:1m 30s

Question 4292Question →

A satellite of mass $m$ moves in a circular orbit around a uniform spherical planet of radius $R$ . If the height of the satellite above the planet's surface is $h = 2R$ and the acceleration due to gravity at the planet's surface is $g$ , which of the following expressions represents the orbital speed of the satellite?

\sqrt{\frac{gR}{3}}

\sqrt{\frac{gR}{2}}

\sqrt{\frac{2gR}{3}}

\sqrt{3gR}

Show answer & explanation

Answer: $\sqrt{\frac{gR}{3}}$

Answer

The orbital speed of the satellite is

\sqrt{\frac{gR}{3}}

.

The total orbital radius from the center of the planet is

r = R + h = R + 2R = 3R

. Since surface gravity is

g = \frac{GM}{R^2}

, we have

GM = gR^2

. Substituting these into the orbital velocity expression

v = \sqrt{\frac{GM}{r}}

yields

v = \sqrt{\frac{gR^2}{3R}} = \sqrt{\frac{gR}{3}}

, which makes the expression

\sqrt{\frac{gR}{3}}

correct.

Step-by-Step Solution

1

Determine the total orbital radius from the center of the planet.

r = R + h = R + 2R = 3R

Gravitational attraction and circular orbital radii are always measured from the center of mass of the primary body, not its surface.

2

Relate the gravitational constant

G

and planet mass

M

to surface gravity

g

.

g = \frac{GM}{R^2} \implies GM = gR^2

At the surface of a spherical planet of radius

R

, the gravitational field strength is

g

.

3

Substitute

r = 3R

and

GM = gR^2

into the circular orbital velocity equation

v = \sqrt{\frac{GM}{r}}

.

v = \sqrt{\frac{gR^2}{3R}} = \sqrt{\frac{gR}{3}}

Equating centripetal force to gravitational force

\frac{m v^2}{r} = \frac{G M m}{r^2}

yields

v = \sqrt{\frac{GM}{r}}

.

Key Concept

Orbital velocity of a satellite in terms of surface gravitational acceleration and orbital radius

Question 4293Question →

A motorist traveling along a straight highway at a constant speed of $30\text{ m/s}$ observes a road hazard ahead. The driver experiences a reaction delay of $0.5\text{ s}$ before applying the brakes, after which the vehicle decelerates uniformly at a rate of $5\text{ m/s}^2$ . What is the total distance traveled by the vehicle from the instant the hazard is observed until the vehicle comes to a complete stop?

90\text{ m}

97.5\text{ m}

105\text{ m}

195\text{ m}

Show answer & explanation

Answer: $105\text{ m}$

Answer

The total distance traveled by the vehicle before coming to a stop is

105\text{ m}

.

The motion consists of two distinct stages: a constant-speed phase during the

0.5\text{ s}

reaction time (

15\text{ m}

) and a uniformly decelerating phase until rest (

90\text{ m}

). Summing both components yields

105\text{ m}

.

Step-by-Step Solution

1

Calculate the reaction distance (

s_1

) covered during the driver's reaction delay.

s_1 = u \times t_r = 30\text{ m/s} \times 0.5\text{ s} = 15\text{ m}

Before the brakes are applied, the vehicle continues moving at its initial constant speed of

30\text{ m/s}

.

2

Calculate the braking distance (

s_2

) using the third equation of motion.

v^2 = u^2 + 2as_2 \implies 0^2 = 30^2 + 2(-5)s_2 \implies 10s_2 = 900 \implies s_2 = 90\text{ m}

The vehicle decelerates from

30\text{ m/s}

to a final velocity of

0\text{ m/s}

at a uniform deceleration rate of

a = -5\text{ m/s}^2

.

3

Sum the reaction distance and the braking distance to find the total stopping distance.

s_{\text{total}} = s_1 + s_2 = 15\text{ m} + 90\text{ m} = 105\text{ m}

The total stopping distance is the sum of the distance covered before braking begins and the distance covered while braking.

Key Concept

Two-phase linear motion combining constant velocity reaction distance and uniform deceleration braking distance.

Question 4294Question →

What is the indefinite integral $\int \left( (3x + 1)^2 - 6\sin(3x) \right) dx$ ?

3x^3 + 3x^2 + x + 2\cos(3x) + C

3x^3 + 3x^2 + x - 2\cos(3x) + C

3x^3 + 3x^2 + x + 2\cos(3x)

3x^3 + 3x^2 + x + 6\cos(3x) + C

Show answer & explanation

Answer: $3x^3 + 3x^2 + x + 2\cos(3x) + C$

Answer

3x^3 + 3x^2 + x + 2\cos(3x) + C

Expanding

(3x+1)^2

yields

9x^2 + 6x + 1

. Integrating

9x^2 + 6x + 1

gives

3x^3 + 3x^2 + x

. Integrating

-6\sin(3x)

gives

+2\cos(3x)

. Combining these results with the mandatory constant of integration

+ C

gives

3x^3 + 3x^2 + x + 2\cos(3x) + C

.

Step-by-Step Solution

1

Expand the squared polynomial binomial term

(3x + 1)^2 = 9x^2 + 6x + 1

Expanding the expression allows for simple term-by-term integration using the power rule.

2

Integrate the polynomial component term by term

\int (9x^2 + 6x + 1) dx = 3x^3 + 3x^2 + x

Apply the power rule

\int x^n dx = \frac{x^{n+1}}{n+1}

to each polynomial term.

3

Integrate the trigonometric component

\int -6\sin(3x) dx = -6 \left(-\frac{1}{3}\cos(3x)\right) = +2\cos(3x)

The integral of

\sin(kx)

is

-\frac{1}{k}\cos(kx)

. Multiplying by

-6

yields

+2\cos(3x)

.

4

Combine terms and add the constant of integration

3x^3 + 3x^2 + x + 2\cos(3x) + C

Indefinite integrals must include an arbitrary constant

+ C

to account for all antiderivatives.

Key Concept

Indefinite Integration of Polynomial and Trigonometric Functions

Question 4295Question →

A man standing at a distance in front of a tall vertical cliff claps his hands and hears the echo after $1.2\text{ s}$ . If the speed of sound in air is $340\text{ m/s}$ , what is the distance between the man and the cliff in meters?

Show answer & explanation

Answer: 204

Answer

The distance between the man and the cliff is

204\text{ m}

.

An echo is formed when sound travels to an obstacle and reflects back. The time taken for the sound to travel to the cliff and back is

1.2\text{ s}

. The total distance covered by sound is

v \times t = 340\text{ m/s} \times 1.2\text{ s} = 408\text{ m}

. Since this distance covers two equal trips (to the cliff and back), the distance to the cliff is

408\text{ m} / 2 = 204\text{ m}

.

Step-by-Step Solution

1

State the relationship between sound speed, total echo time, and distance.

Total distance traveled by the sound is twice the distance to the cliff:

2d = v \times t

.

An echo involves sound traveling from the source to the reflecting barrier and back.

2

Substitute the given values into the equation.

2d = 340\text{ m/s} \times 1.2\text{ s} = 408\text{ m}

.

To find the total distance traversed by the sound wave.

3

Solve for the distance

d

.

d = \frac{408\text{ m}}{2} = 204\text{ m}

.

The one-way distance to the cliff is half of the total distance traveled by the echo.

Key Concept

Calculation of echo distance using

d = \frac{v t}{2}

Question 4296Question →

If the expression $\frac{3 - \sqrt{5}}{3 + \sqrt{5}}$ is simplified and written in the form $a + b\sqrt{5}$ , where $a$ and $b$ are rational numbers, what is the numerical value of $a + b$ ?

Show answer & explanation

Answer: 2

Answer

The numerical value of

a + b

is

2

.

To express

\frac{3 - \sqrt{5}}{3 + \sqrt{5}}

in the standard form

a + b\sqrt{5}

, multiply both the numerator and denominator by the conjugate of the denominator, which is

(3 - \sqrt{5})

. The numerator expands to

(3 - \sqrt{5})^2 = 9 - 6\sqrt{5} + 5 = 14 - 6\sqrt{5}

. The denominator becomes

3^2 - (\sqrt{5})^2 = 9 - 5 = 4

. Dividing gives

\frac{14}{4} - \frac{6}{4}\sqrt{5} = \frac{7}{2} - \frac{3}{2}\sqrt{5}

. Hence,

a = \frac{7}{2}

and

b = -\frac{3}{2}

, making

a + b = \frac{7}{2} - \frac{3}{2} = \frac{4}{2} = 2

.

Step-by-Step Solution

1

Multiply numerator and denominator by the conjugate of the denominator

\frac{(3 - \sqrt{5})(3 - \sqrt{5})}{(3 + \sqrt{5})(3 - \sqrt{5})}

To eliminate the surd from the denominator.

2

Expand both the numerator and the denominator

\frac{14 - 6\sqrt{5}}{4}

Using

(x-y)^2 = x^2 - 2xy + y^2

for the numerator and difference of two squares

(x+y)(x-y) = x^2 - y^2

for the denominator.

3

Separate into rational component and radical coefficient

\frac{7}{2} - \frac{3}{2}\sqrt{5}

Simplifying fractions by dividing numerator and denominator by their greatest common divisor.

4

Calculate the sum

a + b

\frac{7}{2} - \frac{3}{2} = 2

Comparing

\frac{7}{2} - \frac{3}{2}\sqrt{5}

with

a + b\sqrt{5}

yields

a = \frac{7}{2}

and

b = -\frac{3}{2}

.

Key Concept

Rationalization of Binomial Denominators using Conjugates

Estimated Time:1m 30s

Question 4297Question →

A metallic sphere weighs $5.0\text{ N}$ in air. When completely immersed in water, its apparent weight is $3.0\text{ N}$ . When completely immersed in an unknown liquid $X$ , its apparent weight is $3.4\text{ N}$ . What is the density of liquid $X$ in $\text{kg/m}^3$ ? (Take the density of water as $1000\text{ kg/m}^3$ and acceleration due to gravity $g = 10\text{ m/s}^2$ ).

Show answer & explanation

Answer: 800

Answer

The density of liquid

X

is

800\text{ kg/m}^3

.

The upthrust in water (

2.0\text{ N}

) gives the volume of the sphere as

2.0 \times 10^{-4}\text{ m}^3

. Using the upthrust in liquid X (

1.6\text{ N}

), the density of liquid X is calculated as

\rho_X = \frac{1.6}{(2.0 \times 10^{-4})(10)} = 800\text{ kg/m}^3

.

Step-by-Step Solution

1

Calculate upthrust in water

U_w = 5.0\text{ N} - 3.0\text{ N} = 2.0\text{ N}

Upthrust equals the loss in weight of the submerged body in water.

2

Determine the volume of the metallic sphere

V = \frac{U_w}{\rho_w g} = \frac{2.0}{1000 \times 10} = 2.0 \times 10^{-4}\text{ m}^3

According to Archimedes' principle, upthrust in water equals the weight of displaced water.

3

Calculate upthrust in liquid X

U_X = 5.0\text{ N} - 3.4\text{ N} = 1.6\text{ N}

Loss of weight in liquid X gives the upthrust exerted by liquid X.

4

Calculate the density of liquid X

\rho_X = \frac{U_X}{V g} = \frac{1.6}{(2.0 \times 10^{-4}) \times 10} = 800\text{ kg/m}^3

Rearranging

U_X = \rho_X V g

allows solving for the unknown fluid density.

Key Concept

Archimedes' Principle and Apparent Weight

Estimated Time:2m 0s

Question 4298Question →

An electric heater rated at $50\text{ W}$ is used to heat a solid block of mass $1.5\text{ kg}$ for $4\text{ minutes}$ . During this period, the temperature of the block increases from $30^\circ\text{C}$ to $70^\circ\text{C}$ . If $20\%$ of the heat energy supplied by the heater is lost to the surroundings, what is the heat capacity of the block?

240\text{ J K}^{-1}

160\text{ J K}^{-1}

300\text{ J K}^{-1}

60\text{ J K}^{-1}

Show answer & explanation

Answer: $240\text{ J K}^{-1}$

Answer

240\text{ J K}^{-1}

The correct answer is

240\text{ J K}^{-1}

. The heater delivers

12{}000\text{ J}

of energy in

4\text{ minutes}

. Accounting for

20\%

environmental heat loss leaves

9{}600\text{ J}

absorbed by the block. Dividing this energy by the

40\text{ K}

temperature rise gives a total heat capacity of

240\text{ J K}^{-1}

.

Step-by-Step Solution

1

Calculate total electrical energy supplied by heater

Q_{\text{supplied}} = P \times t = 50\text{ W} \times (4 \times 60\text{ s}) = 12{}000\text{ J}

Electrical work converted to heat is given by power multiplied by time in seconds.

2

Determine useful heat absorbed by block after accounting for energy loss

Q_{\text{useful}} = (1 - 0.20) \times 12{}000\text{ J} = 0.80 \times 12{}000\text{ J} = 9{}600\text{ J}

Since

20\%

of supplied energy is lost,

80\%

is retained to raise the temperature of the block.

3

Calculate temperature rise

\Delta T = 70^\circ\text{C} - 30^\circ\text{C} = 40\text{ K}

Temperature difference is the final temperature minus the initial temperature.

4

Calculate heat capacity of the block

C = \frac{Q_{\text{useful}}}{\Delta T} = \frac{9{}600\text{ J}}{40\text{ K}} = 240\text{ J K}^{-1}

Heat capacity

C

is defined as total heat absorbed per unit temperature change (

C = \frac{Q}{\Delta T}

).

Key Concept

Heat Capacity (

C = \frac{Q}{\Delta T}

) represents total thermal capacity of a body, whereas Specific Heat Capacity (

c = \frac{Q}{m\Delta T}

) is heat capacity per unit mass.

Question 4299Question →

A student stands between two tall, parallel vertical walls and claps her hands once. She hears the first echo after $1.0\text{ s}$ and the second echo after $1.5\text{ s}$ . Taking the speed of sound in air to be $340\text{ m s}^{-1}$ , what is the distance between the two walls?

425\text{ m}

850\text{ m}

510\text{ m}

255\text{ m}

Show answer & explanation

Answer: $425\text{ m}$

Answer

The distance between the two walls is

425\text{ m}

.

Sound travels from the student to each wall and reflects back. The distance to the first wall is

\frac{340 \times 1.0}{2} = 170\text{ m}

and to the second wall is

\frac{340 \times 1.5}{2} = 255\text{ m}

. Adding both distances gives the total separation between the walls as

425\text{ m}

.

Step-by-Step Solution

1

Calculate the distance from the student to the closer wall (

d_1

)

d_1 = \frac{v \times t_1}{2} = \frac{340 \text{ m s}^{-1} \times 1.0 \text{ s}}{2} = 170\text{ m}

Sound travels to the wall and reflects back, so the time given corresponds to twice the distance.

2

Calculate the distance from the student to the further wall (

d_2

)

d_2 = \frac{v \times t_2}{2} = \frac{340 \text{ m s}^{-1} \times 1.5 \text{ s}}{2} = 255\text{ m}

The sound for the second echo travels to the second wall and back.

3

Determine the total distance between the two parallel walls

D = d_1 + d_2 = 170\text{ m} + 255\text{ m} = 425\text{ m}

Since the student is positioned between the two walls, the separation of the walls is the sum of both individual distances.

Key Concept

Echo distance relation

2d = v t

for sound reflection from barriers.

Estimated Time:1m 30s

Question 4300Question →

A metal rod has a linear expansivity of $1.5 \times 10^{-5}\text{ K}^{-1}$ . What is the volume expansivity of a sphere made from the same metal?

1.5 \times 10^{-5}\text{ K}^{-1}

3.0 \times 10^{-5}\text{ K}^{-1}

4.5 \times 10^{-5}\text{ K}^{-1}

6.0 \times 10^{-5}\text{ K}^{-1}

Show answer & explanation

Answer: $4.5 \times 10^{-5}\text{ K}^{-1}$

Answer

The volume expansivity of the sphere is

4.5 \times 10^{-5}\text{ K}^{-1}

.

The volume expansivity

\gamma

of a uniform solid object is related to its linear expansivity

\alpha

by

\gamma = 3\alpha

. Multiplying

1.5 \times 10^{-5}\text{ K}^{-1}

by

3

gives

4.5 \times 10^{-5}\text{ K}^{-1}

.

Step-by-Step Solution

1

Identify the mathematical relationship between linear expansivity (

\alpha

) and volume (cubical) expansivity (

\gamma

).

\gamma = 3\alpha

For an isotropic solid, volume expansion occurs equally in three dimensions, making the volume coefficient three times the linear coefficient.

2

Substitute the given value of linear expansivity into the formula and calculate.

\gamma = 3 \times (1.5 \times 10^{-5}\text{ K}^{-1}) = 4.5 \times 10^{-5}\text{ K}^{-1}

Obtain the numeric value of volume expansivity.

Key Concept

Relationship between linear and volume expansivity of solids

Answer

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