Question

the velocity-displacement graph for a particle on straight runway is shown in figure. the acceleration of a particle at S=150m is
1• 125/8 m/s
2•125/4m/s
3•250/4m/s
4•60m/s​

Answer 1

125/8ms ...................

Answer 2

Answer:

125/8 m/s^2.

Explanation:

The graph changes at the velocity of 50m/s. So, the journey must be taken from the 100 to 200 m. So, the distance will be 200-100=100 m. THe final velocity of the particle on the straight runway will be given as 75 m/s and the initial velocity of the particle on the straight runway is 50 m/s.

So, on applying v^2=u^2+2aS. Where S is the distance, a is the acceleration of the particle, v is the final velocity and u is the initial velocity. So, on substituting the values on the equation we will get 75^2-50^2=2*a*100 which on solving we will get 3125/2*100 =a so, the acceleration will be 125/8 m/s^2.