Two cars start from rest at a red stop light. When the light turns green, both c

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 m/s2 for 4.9 seconds. It then continues at a constant speed for 11.4 seconds, before applying the brakes such that the car’s speed decreases uniformly coming to rest 306.03 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 fast is the blue car going 1.5 seconds after it starts? m/s

2)How fast is the blue car going 10.4 seconds after it starts? m/s

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

4) What is the acceleration of the blue car once the brakes are applied? m/s2

5) What is the total time the blue car is moving? s

6) What is the acceleration of the yellow car?

Explanation / Answer

1)

V=U+at

when car starts intial velocity equals to zero

blue car velocity after it starts at t=1.5 seconds

V=at=4*1.5=6m/s

2)

After 10.4s s the blue car is going at the speed it reached after 4.9 s , which was 4*4.9= = 19.6 m/s

V=4*4.9=19.6m/s

3)

Taking the blue car into consideration, you have 3 distinct distances: D1 - car accelerating, D2, car cruising, D3 - Decelerating.

totaldistance=306.03m

D1 accelrating

s=0.5at2=0.5* 4*4.92=48.02m

D2 blue car is moving constant speed=19.6*11.4=223.44m

blue car travel before 48.02+223.44=271.46m

total distance=306.03m

distance travelled by car during decelration

D3=306.03-271.46

=34.57m

4)

v² = u² + 2as
· v = 0, u = 19.6, s = 34.57

0 = 19.6² + 2a(34.57)
69.14a = - 384.16
a = -5.55 m/s² =====> Acceleration applied by blue car when brakes applied

5)

, you need to use the v = u + at formula, and you will get

t3 = (v-u)/a = (0 - 19.6) / (- 5.55) = 3.53s

total time taken by blue car=4.9+11.4+3.53

=19.83s

6)

For the last part of the yellow car, you will need to use:
s = ut + ½at², u=0, t = 19.83, s =306.03

306.03= (0.5)(19.83)² a =196.61 a

accelration of yellow car=1.55m/s2

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