Question

The speed-time graph of a particle moving along a fixed direction is shown in Fig. 3.28. Obtain the distance traversed by the particle between (a) t = 0 s to 10 s, (b) t = 2 s to 6 s.

(Fig. 3.28)

What is the average speed of the particle over the intervals in (a) and (b)?

Answer 1

(a) Distance travelled by the particle = Area under the given graph

Average speed =

(b) Let s₁ and s₂ be the distances covered by the particle between time

t = 2 s to 5 s and t = 5 s to 6 s respectively.

Total distance (s) covered by the particle in time t = 2 s to 6 s

s = s₁ + s₂ … (i)

For distance s₁:

Let u′ be the velocity of the particle after 2 s and a′ be the acceleration of the particle in t = 0 to t = 5 s.

Since the particle undergoes uniform acceleration in the interval t = 0 to t = 5 s, from first equation of motion, acceleration can be obtained as:

v = u + at

Where,

v = Final velocity of the particle

12 = 0 + a′ × 5

Again, from first equation of motion, we have

v = u + at

= 0 + 2.4 × 2 = 4.8 m/s

Distance travelled by the particle between time 2 s and 5 s i.e., in 3 s

For distance s₂:

Let a″ be the acceleration of the particle between time t = 5 s and t = 10 s.

From first equation of motion,

v = u + at (where v = 0 as the particle finally comes to rest)

0 = 12 + a″ × 5

Distance travelled by the particle in 1s (i.e., between t = 5 s and t = 6 s)

From equations (i), (ii), and (iii), we get

s = 25.2 + 10.8 = 36 m