Question

The distance ($m$) vs. time ($t$) graph of a particle at $t=0$ makes an angle of ${45}^0$ with time axis. After $1$ second it makes an angle of ${60}^0$ with time axis, what is the acceleration (in $m/s^2$) of the particle, assuming its acceleration is constant for given time period?

• A. $\sqrt{3}-1$
• B. $\sqrt{2}+1$
• C. $\sqrt{3}$
• D. 1

Answer 1

The correct option is A $\sqrt{3}-1$

$v_0=\frac{s_0}{t_1}=\tan {45}^{\circ }=1$
$v_1=\frac{s_1}{t_1}=\tan {60}^{\circ }=\sqrt{3}$
$a=\frac{v_1-v_0}{1}$
$=\frac{\sqrt{3}-1}{1}$
$=\sqrt{3}-1m/s^2$