A particle starting from rest moves in a straight line with acceleration shown in graph. Find distance(in m) travelled by particle in first four seconds of its motion.
 

Dear Student,

The above graph is case of variable acceleration.

so lets take this time dependent acceleration to be= a(t)

from graph  we have

a(t)=1-t   (  a straight line with negative slope )

$$\begin{gathered} we have condition \\ for t=0 v\left(t\right)=0 and s\left(t\right)=0 \\ v\left(t\right)=\int a\left(t\right)dt\Rightarrow t-\frac{t^2}{2}+const \\ \sin ce for t=0 v\left(t\right)=0 or, const=0 \\ s\left(t\right)=\int v\left(t\right)dt\Rightarrow \frac{t^2}{2}-\frac{t^3}{6}+const \\ \sin ce s\left(t\right)=0 for t= 0 we have const=0 \\ s(t=1)=\frac{1}{2}-\frac{1}{6}\Rightarrow =0.33 m d(t=1)=0.33 \\ s(t=2)=\frac{4}{2}-\frac{8}{6}=0.67 m d(t=2)=0.34 m \\ s(t=3)=\frac{9}{2}-\frac{27}{6}=0 m d(t=3)=0.67 m \\ s(t=4)=\frac{16}{2}-\frac{64}{6}=-2.67 m or after 4 sec d(t=4)=2.67 m \\ Total Dis\tan ce travelled in 4 sec=4.01 m \end{gathered}$$

Regards.