Car A, traveling at a constant speed of 20\text{ m/s} along a straight horizontal road, pa… — JAMB UTME

Car $A$ , traveling at a constant speed of $20\text{ m/s}$ along a straight horizontal road, passes a landmark $50\text{ m}$ ahead of car $B$ , which is initially stationary. At $t = 0\text{ s}$ , car $B$ starts moving in the same direction, accelerating uniformly at $3.0\text{ m/s}^2$ until it reaches a top speed of $30\text{ m/s}$ , after which it continues at this constant top speed. How many seconds after $t = 0\text{ s}$ does car $B$ catch up with car $A$ ?

Cevap: 20 s

Cevap

Car B catches up with car A after 20 seconds.

Car B accelerates for

10\text{ s}

covering

150\text{ m}

to reach

30\text{ m/s}

. During these

10\text{ s}

, car A reaches a position of

250\text{ m}

(taking into account its

50\text{ m}

head start). Car B then closes the remaining

100\text{ m}

gap at a relative speed of

10\text{ m/s}

(

30\text{ m/s} - 20\text{ m/s}

), taking an extra

10\text{ s}

. The total elapsed time is

10\text{ s} + 10\text{ s} = 20\text{ s}

Adım Adım Çözüm

Calculate the duration

t_1

of car B's acceleration phase to reach

30\text{ m/s}

t_1 = \frac{v_{max} - u}{a} = \frac{30 - 0}{3.0} = 10\text{ s}

Car B accelerates uniformly from rest at

3.0\text{ m/s}^2

until reaching its top speed limit.

Determine the distance

s_B

covered by car B and the position

s_A

of car A at

t = 10\text{ s}

s_B = \frac{1}{2}a t_1^2 = \frac{1}{2}(3.0)(10)^2 = 150\text{ m}

;

s_A = 50 + v_A t_1 = 50 + (20)(10) = 250\text{ m}

Car A starts

50\text{ m}

ahead and moves continuously at

20\text{ m/s}

Find the separation distance between the two cars at

t = 10\text{ s}

and compute the time

\Delta t

required to close it.

\text{Separation} = 250 - 150 = 100\text{ m}; \Delta t = \frac{100}{30 - 20} = 10\text{ s}

Beyond

t = 10\text{ s}

, car B travels at a constant relative velocity of

10\text{ m/s}

faster than car A.

Sum the acceleration time and constant speed time to find the total time taken.

t_{total} = t_1 + \Delta t = 10\text{ s} + 10\text{ s} = 20\text{ s}

Combining both phases yields the exact instant car B overtakes car A.

Anahtar Kavram

Multi-stage relative motion with acceleration limits

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