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4.1 , newtons laws of motion — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysics4.1 , newtons laws of motion1 Mark
MCQ
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
  • ✓
    $v_0+\frac{\pi F_0 T_0}{4 m}$
  • B
    $v_0+\frac{\pi F_0}{2 m}$
  • C
    $v_0+\frac{\pi T_0^2}{4 m}$
  • D
    $v_0+\frac{\pi F_0 T_0}{m}$
✓

Answer

Correct option: A.
$v_0+\frac{\pi F_0 T_0}{4 m}$
a
(a)

$\int F d t=m \Delta v$

$\int F d t$ is area under $F-t$ curve

$m \Delta v \left.=\pi\left(\frac{F_0}{2}\right) \cdot\left(\frac{T_0}{2}\right) \text { [area }=\frac{\pi a b}{2}\right]$

$v-v_0 =\frac{\pi F_0 T_0}{4 m}$

$v=v_0+\frac{\pi F_0 T_0}{4 m}$

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