Question

Comment on the type of motion represented by the velocity-time graph. What conclusion can you draw from this graph?

Answer 1

1. The body is at rest from $t=0s$ to $t=2s$.
2. After $t=2s$,the curve has a positive slope. This means that the motion is accelerated.
3. The slope of the velocity-time graph gives the value of acceleration.
4. Mathematically, the slope of a curve containing points $(x_1,y_1)and(x_2,y_2)$ is given by the equation
   $slope=\frac{y_2-y_1}{x_2-x_1}$
5. Consider the slope of the graph given below.
   
   $$\begin{gathered} slope=\frac{30-0}{6-2} \\ \Rightarrow slope=7.5 \end{gathered}$$
6. Thus, the acceleration of the body is $7.5m/s^2$

Hence, the body is at rest from $t=0s$ to $t=2s$, then moves with a uniform acceleration of $7.5m/s^2$ from $t=2sto6s$.