Question

A body moving with a constant acceleration Travels the distances 3 metre and 8 metre respectively in 1 seconds and 2 seconds calculate the initial velocity and acceleration of the body ​

Answer 1

Answer:

s1= 3m

t1=1s

V1= s1/t1= 3m/s

s2= 8m

t2= 2s

V2=s2/t2=4m/s

therefore,u=3m/s,V=4m/s

a= ∆v/∆t

=4-3/2-1=1m/s²

Answer 2

Answer:-

a = 5 m/s²

u = 0.5 m/s

Given:-

Distance cover in 1s = 3m

Distance cover in 2s = 8m

To find:-

It's initial velocity and acceleration.

Solution:-

Let it's initial velocity be u and acceleration be a.

By using 4th equation of motion:-

$\huge{\boxed{S_n =u +\dfrac{a}{2}(2n-1)}}$

→$\mathsf{S_1 = u + \dfrac{a}{2}(2\times 1 -1)}$

→$\mathsf{3 = u + \dfrac{a}{2}\times (2-1)}$

→$\mathsf{3 = u + \dfrac{a}{2}}$

→$\mathsf{3 = \dfrac{2u + a}{2}}$

→$\mathsf{6 = 2u + a}-------1)$

→$\mathsf{ S_2 = u + \dfrac{a}{2}(2n-1)}$

→$\mathsf{8 = u + \dfrac{a}{2}(2\times 2 -1)}$

→$\mathsf{ 8 = u + \dfrac{a}{2}(4-1)}$

→$\mathsf{8 = u + \dfrac{a}{2}\times 3}$

→$\mathsf{8 = u + \dfrac{3a}{2}}$

→$\mathsf{8 = \dfrac{2u + 3a}{2}}$

→$\mathsf{16 = 2u + 3a}------2)$

Subtract eq. 1 and 2.

→$\mathsf{2u + a -2u -3a = 6-16}$

→$\mathsf{-2a = -10}$

→$\mathsf{a = \dfrac{-10}{-2}}$

→$\mathsf{a = 5 m/s^2}$

Put the value of a in eq. 1

→2u + a = 6

→2u + 5 = 6

→2u = 6 - 5

→2u = 1

→u = 1/2

→u = 0.5 m/s.