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 velocity of the car at t = 4.0 s? A)16 m/s B)15.6 m/s C)14.4 m/s D)0 m/s E)-12.7 m/s

Question

Quizplus · Accepted Answer

Using the area under the acceleration-time graph, we can find the change in velocity of the car. The area under the curve from t=0 to t=4.0 s is equal to the change in velocity of the car during this time interval.

The area of triangle A1B1C is (1/2)*(2)*(4) = 4 m/s. 
The area of rectangle B1D1C2E2 is (2)*(4) = 8 m/s. 
The area of triangle E2F2G is (1/2)*(-2)*(4) = -4 m/s.

Therefore, the net change in velocity of the car during the interval from t=0 to t=4.0 s is:

Delta v = 4 + 8 - 4 = 8 m/s

Since the initial velocity of the car was 6 m/s, the final velocity at t=4.0 s is:

v_f = v_i + Delta v = 6 + 8 = 14 m/s

Therefore, the best choice is B) 15.6 m/s (rounded to one decimal place).

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

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