Question

A particle starts from rest and moves in a straight line with constant acceleration passing two points A and B, which are 100m apart in 15s. If the velocity of the particle at B is three times its velocity at A, find: (a) the velocity of the particle at A; (b) its acceleration; (c) the distance of A from the starting point.

Answer 1

Given:

$v_B=3v_A$

$d=100\;\rm m$

$t=15\:\rm s$

(a)

$$d=\frac{v_B^2+v_A^2}{2a}$$

$$d=\frac{v_B+v_A}{2}t=\frac{3v_A+v_A}{2}t=2v_At$$

$$v_A=\frac{d}{2t}=\frac{100\:\rm m}{2\times 15}=3.33{\: \rm m/s}\\ v_B=3v_A=3\times 3.3=10\:\rm m/s$$

(b)

$$a=\frac{v_B-v_A}{t}=\frac{10-3.33}{15}=0.44\:\rm m/s^2$$

(c)

$$d_0=\frac{v_A^2-v_0^2}{2a}=\frac{3.33^2}{2\times 0.44}=12.6\:\rm m$$