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.What is the average speed of the particle over the intervals in (a) and (b) ?
Answers
whatever the case maybe..
you have a speed-time graph
you can find the distance by finding the area under the curve (made by speed-time graph)
find the area of the shape made by the curve from t=0 to t=10
and then do it for t=2 to t=6
if you need a more specific guidance.. please provide the graph..
Answer:
(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‘ 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/s
