Answer to Question #92503 in Mechanics | Relativity for Deepak

Question

The points A,B and C lie on the same straight line, the distance between AB being 60m. A particle crosses the point A with the velocity 2m/s and moves with uniform acceleration along the straight line. After crossing the point B, it takes 10 sec to reach C with velocity 5m/s. Find the acceleration of the particle.

Answer 1

Let

acceleration of the particle = a

time to travel from A to B = t

speed at point A = V_A

speed at point C = V_C

According to constant acceleration motion formulas

$|AB| = V_A\cdot t+a\frac{t^2}{2}$ $a=\frac{V_C-V_A}{t+10}$ $60=2t+a\frac{t^2}{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$ $a=\frac{5-2}{t+10}=\frac{3}{t+10}\ \ \ (2)$

Substituting a from (2) into equation (1):

$60=2t+\frac{3t^2}{2(t+10)}$ $120(t+10)=4t(t+10)+3t^2$ $120t+1200=4t^2+40t+3t^2$ $7t^2-80t-1200=0$

using quadratic formula:

$t_{1,2}=\frac{-(-40)\pm \sqrt{40^2-7\cdot(-1200)}}{7}=$ $\frac{40\pm \sqrt{10000}}{7}=\frac{40\pm 100}{7}$

$t_1,=20,\ t_2=-60/7$

Since t > 0 t = t₁ = 20s

From equation (2)

$a=\frac{3}{20+10}=\frac{3}{30}=\frac{1}{10}m/s^2$

Answer:

$\frac{1}{10}m/s^2$