Question

the speed time graph of a particle moving along a fixed direction is shown in the figure obtain the distance transversed by the particle between 0 to 10 s and 2 to 5 sec and what is the average speed of the particle over the intervals in a and b
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Answer 1

Answer:

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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: