Question

Two rather unequal cars are competing in a drag race. For simplicity, let's assume that the accelerations of the cars are constant. Car 1 has an acceleration of 3.7 m/s², while Car 2 has an acceleration of 6.3 m/s². Car 2 gets off to a late start, 1.1 seconds after Car 1.
At what time after Car 1 took off will it be overtaken by Car 2? (in s)

A: 3.0

B: 3.8

C: 4.7

D: 5.9

E: 7.4

How far down the track will this happen? (in m)

A: 22

B: 26

C: 30.

D: 35

E: 41

How much faster will Car 2 be than Car 1 by that time? (in m/s)

A: 3.7

B: 4.2

C: 4.7

D: 5.3

E: 6.0

Answer 1

given that

a1 = 3.7 m/s^2

a2 = 6.3 m/s^2

time difference between both the cars = 1.1 s

(a)

distance covered by car 1 in time t = distance covered by car 2 in time (t-1.1)

1/2*3.7*t^2 = 1/2*6.3*(t-1.1)^2

3,7*t^2 = 6.3 * (t^2 + 1.21 -2.2*t)

2.6*t^2 -13.86*t + 7.623= 0

t = 4.70 s or 0.622 s

answer is C : 4.7 s

(b)

d = 1/2*3.7*4.7*4.7 = 40.86 m

answer is E:41 m

(c)

v1 = a1* t = 3.7*4.7 = 17.39 m/s

v2 = a2*(t-1.1) = 6.3*3.6 = 22.68 m/s

v2 - v1 = 22.68 - 17.39 = 5.29 m/s faster than car 2