Question

The speed time graph of a particle moving along a fixed direction is shown in the figure below. Obtain the distance travelled by the particle between (a) t = 0 to t = 10 s (b) t = 2 to 6 s. what is the average speed of the particle over the intervals in (a) and (b)?

Answer 1

Hey mate,

# Answer-
(a) From t=0 to t=10s, s=60m, v=6m/s
(a) From t=2s to t=6s, s=36m, v=9m/s

# Explaination-
(a) 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‘t^2
= 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″t^2
= 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 = Distance / time = 36 / 4 = 9 m/s

Hope that was useful...

Answer 2

(a) 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 ne found by:

v = u + at

Where,

v = Final velocity of the particle

12 = 0 + a′×5

a′ = 12/5 = 2.4 ms-2

v = u + at .....(i)

= 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‘t^2

= 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″t^2

= 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 = Distance / time = 36 / 4 = 9 m/s