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4-1.Newton's Laws of Motion
medium
The magnitude of force acting on a particle moving along $x$-axis varies with time $(t)$ as shown in figure. If at $t=0$ the velocity of particle is $v_0$, then its velocity at $t=T_0$ will be
A
$v_0+\frac{\pi F_0 T_0}{4 m}$
B
$v_0+\frac{\pi F_0}{2 m}$
C
$v_0+\frac{\pi T_0^2}{4 m}$
D
$v_0+\frac{\pi F_0 T_0}{m}$
Solution
(a)
$\int F d t=m \Delta v$
$\int F d t$ is area under $F-t$ curve
$m \Delta v \left.=\pi\left(\frac{F_0}{2}\right) \cdot\left(\frac{T_0}{2}\right) \text { [area }=\frac{\pi a b}{2}\right]$
$v-v_0 =\frac{\pi F_0 T_0}{4 m}$
$v=v_0+\frac{\pi F_0 T_0}{4 m}$
Standard 11
Physics
