Question

A particular moves vertically with an upward initial speed $v_0=10.5m{s}^{-1}$. If its acceleration varies with time as shown in $a-t$ graph in Fig. find the velocity of the particle at $t=4s$.

Answer 1

The velocity of the particle at $t=4s$ can be given as
${\overrightarrow{v}}_4={\overrightarrow{v}}_0+\Delta \overrightarrow{v}$
where $\Delta \overrightarrow{v}\equiv A$(= area under $a-t$) graph during first $4s$)
Referring to $a-t$ graph (shown in the figure), we have
$A=A_1+A_2-A_3-A_4$.........(1)
where

$A_1=5\times 1=5,A_2=\frac{1}{2}\times x\times 5$
$A_3=\frac{1}{2}\times (1-x)\times 10$

$A_4=\frac{1}{2}\times 2\times 10=10$
We can find the value of $x$ as follows:
Using properties of similar triangles, we have $\frac{x}{5}=\frac{1-x}{10}$
This yields $x=\frac{1}{3}$
Substituting $x=\frac{1}{3}$ in $A_2=\frac{1}{2}\times x\times 5$ and $A_3=\frac{1}{2}(1-x)\times 10$,
$A_2=\frac{5}{6}$

and $A_3=\frac{10}{3}$
Then substituting $A_1,A_2,A_3$ and $A_4$ in Eq. (1) we have
$A=-7.5\Rightarrow \Delta v=-7.5$ and $\overrightarrow{v_0}=10.5m/s$
Hence, we have $v_4=v_0+\Delta v=10.5-7.5=3m/s$