A car is stopped at a traffic light. It then travels along a straight road so th

Question

A car is stopped at a traffic light. It then travels along a straight road so that its distance from the light is given by x(t)=bt^2-ct^3, where b= 2.40 m/s^2 and c= 0.130 m/s^3.
1. Calculate the average velocity of the car for the time interval t= 0 to t= 10.0 .
2. Calculate the instantaneous velocity of the car at t=0.
3. Calculate the instantaneous velocity of the car at t=5.00 s.
4. Calculate the instantaneous velocity of the car at t=10.0 s.
5. How long after starting from rest is the car again at rest?

Explanation / Answer

Given Distance x(t) = bt^2 - ct^3 x at t=0,x1 =0 m x at t =10, x2 = 100 b-1000c                     = 100 (2.40)-1000(0.130) = 110 m where b = 2.40 m/s^2, c = 0.130 m/s^2 1.The average velocity is v_avg = 110 m/10s       v_avg =   11 m/s   2.The instantaneous velocity at t=0                         velocity v = dx/dt v = d(bt^2 - ct^3)/dt v = 2bt-3ct^2 v =dx/dt = 2bt-3ct^2 At t=0, v = 0 m/s 3.At t=5s, v = (2*2.40 *5s)-(3 *0.130*25) =14.25 m/s 4.At t=10s, v =(2 *2.40 *10)-(3*0.130*100) = 9m/s 5.The distance travelled in time interval 10s is x(t) = bt^2 - ct^3       = 2.40(100)-0.130(1000) =110m

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