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Zorluk: KolayKinematics and Linear Motion

A car traveling along a straight road at an initial speed of 20 m/s20\text{ m/s} applies its brakes, causing a uniform deceleration of 5 m/s25\text{ m/s}^2 until it comes to a complete stop. What is the total distance traveled by the car during this braking period?

Cevap: 40 m

Cevap

The total distance traveled by the car while coming to a stop is 40 m40\text{ m}.
Using the third equation of linear motion v2=u2+2asv^2 = u^2 + 2as with u=20 m/su = 20\text{ m/s}, v=0 m/sv = 0\text{ m/s}, and acceleration a=5 m/s2a = -5\text{ m/s}^2, we obtain 0=40010s0 = 400 - 10s, which simplifies directly to s=40 ms = 40\text{ m}.

Adım Adım Çözüm

1
Identify the given kinematic parameters from the problem statement.
u=20 m/su = 20\text{ m/s}, v=0 m/sv = 0\text{ m/s}, a=5 m/s2a = -5\text{ m/s}^2
The car decelerates to a stop, so final velocity is zero and acceleration is negative relative to initial direction of motion.
2
Select the appropriate equation of motion linking uu, vv, aa, and displacement ss.
v2=u2+2asv^2 = u^2 + 2as
This equation directly relates initial velocity, final velocity, acceleration, and distance without requiring time.
3
Substitute the known values into the equation and solve for distance ss.
02=202+2(5)s    10s=400    s=40 m0^2 = 20^2 + 2(-5)s \implies 10s = 400 \implies s = 40\text{ m}
Algebraic simplification yields the stopping distance.

Anahtar Kavram

Uniformly Accelerated Motion and Stopping Distance
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