Question

Exercise 2.7-Enhanced-with Solution Constants Part A 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 z(t)t where b-2.50 m/s2 and c 0.100 m/s You may want to review (Pages 37-40) For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of Average Calculate the average velocity of the car for the time interval t 0tot 10.0 s and instantaneous velocities. Submit Part B Calculate the instantaneous velocity of the car at t-0. In Submit Request Answer Part C Calculate the instantaneous velocity of the car at t 5.00 s Submit Request Answer Part D Calculate the instantaneous velocity of the car at t10.0s Submit Request Answer Part E How long after starting from rest is the car again at rest? Submit Request Answer

Answer 1

A.)

Given Distance X(t) = bt^2-ct^3

Distance at t = 0 s,

X(t) = 0 m

Distance at t= 10 s,

X(t=10s) = (2.5m/s^2)*(10s^2)-(0.100 m/s^3)*(10s)^3 = 150 m

Average velocity = total distance / total time taken

= 150m/10s = 15 m/s...Answer.

B.) Instantneous velocity = dx(t)/dt = d(bt^2-ct^3)/dt

= 2bt-3ct^2

at t= 0 s, Vinst. = 2*(2.5m/s^2)*(0s)-3*(0.100m/s^3)*(0s)^2 = 0 m/s.....Answer.

C.)

At t= 5 s, Vinst. = 2*(2.5m/s^2)*(5s)-3*(0.100m/s^3)*(5s)^2

= 17.5 m/s....answer

D.)

At t =10s, Vinst. = 2*(2.5m/s^2)*(10s)-3*(0.100m/s^3)*(10s)^2

= 20 m/s......Answer.

E.)

Velocity is zero when it is in rest.

so, 2bt-3ct^2 =0

> 5t-0.300t^2= 0

> t(5-.300t) = 0

> t = 5/0.300= 16.66s...Answer.

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