Question

Position-time graph of a particle of mass 1 kg is as shown. Change in linear momentum in 16 S is

Answer 1

By the simple line equation apply for above figure

then by, $y=mx+c$

$x=\left(\frac{y_2-y_1}{x_2-x_1}\right)t+c$

$x=\left(\frac{20-0}{5-0}\right)t+c$

$x=4t+c$ ...(1) put $\left(\begin{array}{c}x=20\\ t=5\end{array}\right)$ in (1)

$\frac{20}{5\times 4}=c,c=\frac{20}{20}=1,c=1$

then eq${}^n$ becomes

$x=4t+1$

then, velocity $\left(v\right)=\frac{dx}{dt}=4,v=4m/sec$

then linear momentum, $p=mv,m=1kg$ as given

$p=1\times 4$

$p=4\frac{kg-m}{sec}$