Question

1. The speed-time graph of a particle moving along a fixed direction is shown in figure. Find:
Distance travelled by the particle between 0 sec to 10 sec
Average speed between this interval
The time when the speed was minimum
(iv)
The time when speed was maximum.
Speed (ms)​

Answer 1

Explanation:

Distance travelled by the particle = Area under the given graph

= (1/2) × (10 – 0) × (12 – 0) = 60 m

Average speed = Distance / Time = 60 / 10 = 6 m/s

(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

a′ = 12 / 5 = 2.4 ms-2

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

s1 = u‘ t + (1/2)a‘ t2

= 4.8 × 3 + (1/2) × 2.4 × (3)2

= 25.2 m ……..(ii)

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

a″ = -12 / 5 = – 2.4 ms-2

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

s2 = u“ t + (1/2)a″ t2

= 12 × 1 + (1/2) (-2.4) × (1)2

= 12 – 1.2 = 10.8 m ………(iii)

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

s = 25.2 + 10.8 = 36 m

∴ Average speed = 36 / 4 = 9 m

Answer 2

Answer:

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