1- How fast is the blue car going 3.8 seconds after it starts? 15.2 2-How fast i

Question

1- How fast is the blue car going 3.8 seconds after it starts?

15.2

2-How fast is the blue car going 9.5 seconds after it starts?

19.2

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

257.28

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

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

6-What is the acceleration of the yellow car?

Car Ride 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 4.8 seconds. It then continues at a constant speed for 11 seconds, before applying the brakes such that the car's speed decreases uniformly coming to rest 287.69 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

Explanation / Answer

1)

speed of blue car vb1 = vo + a*t = 0 + (4*3.8) = 15.2 m/s <<<---answer

2)

from 4.8s to 11 s the speed is constant

after time 4.8 speed vb2 = vo + a*t = 0 + (4*4.8) = 19.2 m/s <<<<---answer

3)

x1 = v0*4.8 + (0.5*a*4.8^2) = 0 + (0.5*4*4.8^2) = 46.08 m

x2 = vb2*11 = 19.2*11 = 211.2 m

distance travelled before its brakes are applied to slow down = x1 + x2 = 257.28 m

(4)

for the last part

vf^2 - v2b^2 = 2*a1*(287.69-257.28)

0-19.2^2 = 2*a1*(287.69-257.28)

a1 = -6.06 m/s^2

(5)

time taken for decelartion t3 = vb2/a1 = 19.2/6.06 = 3.2 s

T = t1 + t2 + t3

T = 4.8 + 11 + 3.2 = 19 s

(6)

foryellow car

distance x = 287.69 m

time = T = 19 s

initial velocity = 0

x = vo*T + 0.5*a*T^2

287.69 = 0.5*a*19^2

a = 1.6 m/s^2

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