A car is initially seen to be already traveling at 20 m/s. After 2 seconds at th

Question

A car is initially seen to be already traveling at 20 m/s. After 2 seconds at that speed, it slows with a constant acceleration of 2 m/s^2, and it does that for 3 seconds It then stops accelerating and travels another 28 meters. After that, it slows to a stop over a distance of 14 meter with another constant acceleration. Sketch the acceleration, velocity, and position graphs. Label all points you used to connect the lines. What is a total distance traveled by the car between the time it was initially seen to the time when it stopped? How much total time did that take?

Explanation / Answer

Vi = 20 m/s

after 2sec

travelled distance in 2 sec = 20*2 = 40 m

after 2 sec a = -2 m/s^2

after 3 sec

distance travelled

s = ut+1/2*a*t^2 = 20*1 -0.5*2*1^2 = 19 m

velocity after 3 sec

v= u +at = 20- 2*1 = 18 m/s

distance travelled after 3sec

28 meter

velocity in the end of 28 m = 18 m/s

time taken by this car

v = xt

t = 18/28 =0.6428 sec

after this it stopes in 14 meter

v^2 =u^2 +2*a*s

0 = 18^2 +2*a*14

a = -11.57

time taken in this

v= u +at

0 = 18 -11.57 *t

t = 18 /11.57 = 1.55 sec

so

Total distance =

40+19+28+14 = 101 m

total time = 2+ 1+0.6428 +1.55 = 5.1928 sec

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