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 4.6 m/s2 for 4.4 seconds. It then continues at a constant speed for 8.5 seconds, before applying the brakes such that the car’s speed decreases uniformly coming to rest 247 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. 2)How fast is the blue car going 10.4 seconds after it starts?3)How far does the blue car travel before its brakes are applied to slow down?4)What is the acceleration of the blue car once the brakes are applied?

Answer 1

2) 20.2 m/s

In the first 4.4 seconds of its motion, the blue car accelerates at a rate of

$a=4.6 m/s^2$

So its final velocity after these 4.4 seconds is

$v=u+at$

where

u = 0 is the initial velocity (the car starts from rest)

a is the acceleration

t is the time

Substituting t = 4.4 s, we find

$v=0+(4.6)(4.4)=20.2 m/s$

After this, the car continues at a constant speed for another 8.5 s, so it will keep this velocity until

$t=4.4 + 8.5 = 12.9 s$

Therefore, the velocity of the car 10.4 seconds after it starts will still be 20.2 m/s.

3) 216.2 m

The distance travelled by the car during the first 4.4 s of the motion is given by

$d_1 = ut_1 + \frac{1}{2}at_1^2$

where

u = 0 is the initial velocity

$t_1 = 4.4 s$ is the time

$a=4.6 m/s^2$ is the acceleration

Substituting,

$d_1 = 0 +\frac{1}{2}(4.6)(4.4)^2=44.5 m$

The car then continues for another 8.5 s at a constant speed, so the distance travelled in the second part is

$d_2 = vt_2$

where

$v_2 = 20.2 m/s$ is the new velocity

$t_2 = 8.5 s$ is the time

Substituting,

$d_2 = (20.2)(8.5)=171.7 m$

So the total distance travelled before the brakes are applied is

$d=44.5 m+171.7 m=216.2 m$

4) $-6.62 m/s^2$

We are told that the blue car comes to a spot at a distance of 247 meters from the start. Therefore, the distance travelled by the car while the brakes are applied is

$d_3 = 247 m -216.2 m=30.8 m$

We can find the acceleration of the car during this part by using the SUVAT equation:

$v_f^2 - v_i^2 = 2ad_3$

where

$v_f = 0$ is the final velocity (zero since the car comes to a stop)

$v_i = 20.2 m/s$ is the velocity of the car at the moment the brakes are applied

a is the acceleration

$d_3 = 30.8 m$

Solving for a, we find

$a=\frac{v_f^2 -v_i^2 }{2d}=\frac{0-(20.2)^2}{2(30.8)}=-6.62 m/s^2$