Question

The velocity-displacement graph for a particle on a
straight runway is as shown in figure. The
acceleration of particle at S = 150 m is
​

Answer 1

Answer: It is the problem of uniformly accelerated motion.

Look in the displacement interval 100 to 200.

v2 - u2 = 2aS

Explanation:

Answer 2

Answer:

Required acceleration is 125/8 m/s²

Explanation:

From the graph,

${v}^{2} = {u}^{2} + 2as$

Where v is final velocity, u is initial velocity, a is acceleration and s is displacement

Now,

${v}^{2} - {u}^{2} = 2as$

Hete S= 200-100=100 m

and u= 50 m/s and v= 75 m/s

${75}^{2} - {50}^{2} = 2 \times a \times 100$

So,

$\frac{5625 - 2500}{2 \times 100} = a$

$a = \frac{125}{8} m {s}^{ - 2}$