Question

a particle along a straight line path ABC with the uniform acccelartion 0.5m/s^2 while it crosses a its velocity is 5m/s. it reaches C with the velocity 40m/s, after 30 sec it crossesed B in its path . find distance AB

Answer 1

Answer:

it may be 375m if path is A-B-C

Answer 2

ANSWER:

The distance of AB is 217.5 m

EXPLANATION:

The particle along a straight line path ABC with the uniform acceleration $0.5 m / \mathrm{sec}^{2}$

The new velocity reaches to 5 m/sec

The next velocity mark at point C is 40 m/sec

The object reaches the point C after it crosses the point B after 30 seconds.

The time taken to cover the particle from B to C is 30 seconds

The initial velocity at A is 5 m/sec and the final velocity at C is 40 m/sec

The distance covered from B to C is  

$v^{2}=u^{2}+2 a S$

$40^{2}=5^{2}+2 \times 5 \times S$

$1600=25+10 S$

$157.5 m=S$

Hence, to find the distance from A to C, we use

$S=5 .(30)+\frac{1}{2}(0.5)(30)^{2}$

$S=150+\frac{1}{2}(0.5)(30)^{2}$

$S=375 \mathrm{m}$

Therefore, the distance of A to B is 375 m - 157.5 m = 217.5 m