Acceleration-time graphs in 1D kinematics

A body which is initially at rest experiences a change in velocity and travels in a straight horizontal line as shown below (graph drawn to scale, take positive v as rightwards motion).

Which of the following shows the correct graph for the acceleration of the body over the same time period, and the sign of the displacement of the body after 5 s?

Acceleration Graph A

Acceleration Graph B

Acceleration Graph C

Acceleration Graph B, positive displacement Acceleration Graph C, positive displacement Acceleration Graph A, positive displacement Acceleration Graph B, negative displacement

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1 solution

Wee Xian Bin
Jul 2, 2016

Since d v d t = a \frac{dv}{dt}=a , the acceleration at a time point in the acceleration-time graph is the gradient at the corresponding point in the velocity-time graph. Since velocity is constant between t = 3 t=3 and t = 4 t=4 , acceleration is zero, hence graph B is wrong. Since the gradient of the velocity-time graph between t = 4 t=4 and t = 5 t=5 is smaller than the maximum gradient of the velocity-time graph between t = 0 t=0 and t = 2 t=2 , Graph C is correct and Graph A is wrong.

v d t = s \int v dt = s . Since the area above the x x -axis is greater than the area below the x x -axis in the velocity-time graph, displacement from origin is positive.

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