Question

a.Just as car A is starting up, it is passed by car B. Car B travels with a constant velocity of 15 m/s, while car A accelerates with a constant acceleration of 5.0 m/s², starting from rest.

What distance is traveled by car A in 5 seconds?

b.

Just as car A is starting up, it is passed by car B. Car B travels with a constant velocity of 15 m/s, while car A accelerates with a constant acceleration of 5.0 m/s², starting from rest.

What distance is traveled by Car B in 5 seconds.

c.

Just as car A is starting up, it is passed by car B. Car B travels with a constant velocity of 15 m/s, while car A accelerates with a constant acceleration of 5.0 m/s², starting from rest.

At what time does car A overtake car B?

Answer 1

Part A.

Initial speed of car A = 0 m/sec

acceleration = 5.0 m/sec^2

Using 2nd kinematic equation:

d = U*t + (1/2)*a*t^2

d = 0*5 + (1/2)*5.0*5.0^2

d = 62.5 = distance traveled by car A in 5.0 sec

Part B.

Sine Car B is traveling at constant speed, So

distance = Speed*time = V*t

distance = (15 m/sec)*5.0 sec

d1 = 75 = distance traveled by car B in 5.0 sec

Part C.

Suppose car A overtakes car B at time 't1' then

distance traveled by car A in t1 time = distance traveled by car B in t1 time

U*t1 + (1/2)*a*t1^2 = V*t1

0*t1 + (1/2)*5.0*t1^2 = 15*t1

t1 = 15*2/5.0 = 6 sec

So car A will overtake Car B in 6 sec

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