chapter-header

Selina Solutions Class 9 Physics Solutions for Exercise 2C in Chapter 2 - Chapter 2 Motion In One Dimension

Question 17 Exercise 2C

A body moving with a constant acceleration travels the distances 3m and 8m respectively in 1s and 2s. Calculate: (i) the initial velocity, and (ii) the acceleration of the body.

Answer:

Let distance travelled in time t1=1s be s1=3m

Let distance travelled in time t2=2s be s2=8m

We know that s = ut+\frac{1}{2}at^2

, substituting for s1 and s2 is:

S1= ut_1+\frac{1}{2}at_1^2

and S2= ut_1+\frac{1}{2}at_2^2

Subtracting s1 from s2, we get;

S2-S1 = u(t2-t1) + 1/2a (t2 x t2 – t1 x t1)

8-3 = u(2-1) + 1/2 a(4-1)

5 = u + 3/2a

a = (10-2u)/3 - equation 1

use this value of ‘a’ obtained in the equation for S1, we get;

S1= ut_1+\frac{1}{2}at_1^2

3=u(1 x 1) + ½ [(10-2u)/3](1 x 1)

3 = u + 10−2𝑢 / 6

18= 6u + 10 – 2u

u =2m/s

using the value of u obtained in equation 1, we get;

a= (10-2u)/3

a=6/3

a=2 m/s^2

Connect with us on social media!
2022 © Quality Tutorials Pvt Ltd All rights reserved