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 5.1 m/s2 for 4.4 seconds. It then continues at a constant speed for 12.4 seconds, before applying the brakes such that the car’s speed decreases uniformly coming to rest 387.36 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.

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

What is the acceleration of the yellow car??????????

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

Explanation / Answer

for blue car,
Let us calculate the speed before it applies brake
Speed is same as that at timr t = 4.4 s
vf = vi+a*t
=0 + 5.1*4.4
= 22.44 m/s

let us now calculate the distance through which it moved after brake was applied
for t= 0 to 4.4 s,
d1 = vi*t + 0.5*a*t^2
= 0*4.4 + 0.5*5.1*4.4^2
=49.37 m

for next 12.4 s
a = 0 and it moves with velocity of 22.44 m/s
d2 = v*t
= 22.44*(12.4)
= 278.26 m

distance travelled while decelerating,
d3 = 387.36 - 49.37 - 278.26
= 59.76 m

Let us now calculate the acceleration,
vf^2 = vi^2 + 2*a*d3
0 = 22.44^2 + 2*a*59.76
a = -4.21 m/s^2
Answer: -4.21 m/s^2

-------------------------------------------------------
Lets calculate time taken by blue car while decelerating
vf = vi+a*t3
0 = 22.44 + (-4.21)*t3
t3 = 5.3 s

Total time for blue car = 4.4+12.4+ 5.3 = 22.1 s

for yellow car,
t = 22.1 s
d = 387.36
use:
d = 0.5*a*t^2
387.36 = 0.5*a*(22.1)^2
a = 1.59 m/s^2
Answer: 1.59 m/s^2

------------------------------
Total time for blue and yellow car both is same
Answer: 22.1 s

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