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)? 11th physics , chapter-3

Answer 1

(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

Answer 2

Answer:

( a ) Distance covered by the particle = Area of the given graph

= (1/2)base x height

= (1/2) x (10) x (12) = 60m

Average speed of the particle = 60/10 = 6 m/s

( b )  The distance traversed by the particle between

t = 1s to 8s

let distance travelled in 1 to 5s be S1 and distance travelled from 6 to 8s be S2.

Thus, total distance travelled, S ( in t = 1 to 8 s) = S1 + S2  . . . . . . ( 1 )

Now, For S1.

Let u’ be the velocity of the particle after 1 second  and a’ be the acceleration in the particle from t = 0 to 5s

We know that the particle is under uniform acceleration from  t = 0 to 5s thus, we can obtain  acceleration using the first equation of motion.

v = u + at

where, v = final velocity

12 = 0 + a’(5)

a’ = 2.4 m/s2

Now to find the velocity of the particle at 1s

v = 0 + 2.4 (1)

v =  2.4 m/s = u’ at t = 1s

Thus, the distance covered by the particle in 4 seconds  i.e., from t = 1 to 5 s.

S1 = u’t  +  ½ a’t2

=  2.4 x 4  +  ½ x 2.4 x 42

= 9.6 + 19.2 =28.8 m

Now, for S2

Let a’’ be the uniform acceleration in the particle from 5s to 10s

Using the first law of motion

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

0 = 12 + a’’ x 5

a’’ = -2.4 m/s

Thus, distance travelled by the particle in 3 seconds i.e., between 5s to 8s

S2 = u’’t + ½ a’’t

S2 = 12 x 3 + ½ x(-2.4) x 32

=  36 + (-1.2)x9

S2 = 25.2m

Thus, putting the values of S1 and S2 in equation ( 1 ), we get:

S = 28.8 + 25.2 = 54m

Therefore, average speed = 54 / 7 = 7.71 m/s.

Explanation: