Question

Two cars start from rest at a red stop light. When the light turns green, both cars accelerate forward. The blue car accelerates uniformly at a rate of 5.1 m/s2 for 4.3 seconds. It then continues at a constant speed for 11.9 seconds, before applying the brakes such that the car’s speed decreases uniformly coming to rest 361.87 meters from where it started. The yellow car accelerates uniformly for the entire distance, finally catching the blue car just as the blue car comes to a stop.

1) How far does the blue car travel before its brakes are applied to slow down?

2) What is the acceleration of the blue car once the brakes are applied?

3) What is the total time the blue car is moving?

4) What is the acceleration of the yellow car?

Answer 1

let

vob = 0

velocity of blue car after 4.3s, vb = vob + ab*t

= 0 + 5.1*4.3

= 21.93 m/s

let t1 = 4.3 s
t2 = 11.9 s

1) distance travelled by the blue car before brakes are applied, d = vob*t1 + (1/2)*ab*t1^2 + vb*t2

= 0 + (1/2)*5.1*4.3^2 + 21.93*11.9

= 308.12 m

2) let a is the acceleration of the blue car during brakes are applied.

use, v^2 - u^2 = 2*a*s (u and v are initial and final velocities)

0^2 - 21.93^2 = 2*a*(361.87 - 308.12)

==> a = (0^2 - 21.93^2)/(2*(361.87 - 308.12))

= -4.47 m/s^2

3) time taken to stop, t = (v - u)/a

= (0 - 21.93)/(-4.47)

= 4.91 s

the total time the blue car is moving = t1 + t2 + t

= 4.3 + 11.9 + 4.91

= 21.1 s

4) let ay is the acceleration of the yellow car.

we know, voy = 0

use, d = voy*t + (1/2)*ay*t^2

361.87 = 0 + (1/2)*ay*21.1^2

==> ay = 2*361.87/21.1^2

= 1.63 m/s^2