Worked example 2.1: Velocity-time graph

Question: Consider the motion of the object whose velocity-time graph is given in the diagram.

1. What is the acceleration of the object between times $t=0$ and $t=2$?
2. What is the acceleration of the object between times $t=10$ and $t=12$?
3. What is the net displacement of the object between times $t=0$ and $t=16$?

 
Answer:

1. The $v$-$t$ graph is a straight-line between $t=0$ and $t=2$, indicating constant acceleration during this time period. Hence,
   
   $$a = \frac{{\mit\Delta}v}{{\mit\Delta} t} = \frac{v(t=2)-v(t=0)}{2-0} = \frac{8-0}{2} = 4 {\rm m s^{-2}}.$$
   
   

2. The $v$-$t$ graph is a straight-line between $t=10$ and $t=12$, indicating constant acceleration during this time period. Hence,
   
   $$a = \frac{{\mit\Delta}v}{{\mit\Delta} t} = \frac{v(t=12)-v(t=10)}{12-10} = \frac{4-8}{2} = -2 {\rm m s^{-2}}.$$
   
   
   The negative sign indicates that the object is decelerating.
3. Now, $v=dx/dt$, so
   
   $$x(16)-x(0) = \int_0^{16} v(t) dt.$$
   
   
   In other words, the net displacement between times $t=0$ and $t=16$ equals the area under the $v$-$t$ curve, evaluated between these two times. Recalling that the area of a triangle is half its width times its height, the number of grid-squares under the $v$-$t$ curve is 25. The area of each grid-square is $2\times 2=4 {\rm m}$. Hence,
   
   $$x(16)-x(0) = 4\times 25 = 100 {\rm m}.$$