A car moving along the x direction has acceleration,a_x that varies with time as shown in the figure.At the moment,t = 0 s,the car is located at x = 12 m and has a velocity of 6 m/s in the positive x direction.What is the displacement of the car between t = 1.0 s and t = 2.0 s? A)-0.8 m B)5.6 m C)11.1 m D)2.8 m E)9.5 m

Question

Quizplus · Accepted Answer

First, we need to find the velocity of the car at t=1s:
v= v0 + at  = 6 + (-3) = 3 m/s (as shown in the graph)
Then, we can find the displacement between t=1s and t=2s using:
Δx = v0Δt + 1/2at^2 = 6(1) + 1/2(-3)(1)^2 = 3 m
Finally, we can find the total displacement by subtracting the initial displacement (12m):
Δx_total = Δx2-1 + Δx1 = 3 + (-12) = -9 m
Note that the negative sign means that the car moved in the negative direction, which is consistent with the graph.

A Car Moving Along the X Direction Has Acceleration,ax That

A Car Moving Along the X Direction Has Acceleration,a_x That