Question

If a body covers 26 m and 30 m in the 6th and 7th seconds of its motion, then the initial velocity and acceleration of the body respectively are (assuming acceleration to be uniform

Answer 1

Displacement 6th sec=u+a/2(2×6-1)

=u+11a/2. (1)

Displacement 7th sec=u+a/2(2×7-1)

=u+13a/2. (2)

Subtract (1)&(2)

11a/2-13a/2

=a=4m/s²

Substitute a in (1)

u+11×4/2=26

u+22=26

u=4

acceleration =4

Initial velocity =4

Answer 2

Given :

A body covers 26 m and 30 m in the 6th and 7th seconds of its motion .

To Find :

The initial velocity and acceleration of the body .

Solution :

We know , displacement in nth second is given by :

( assuming acceleration to be uniform )

$s_n=u+\dfrac{1}{2}a(2n-1)$

For , n = 6 :

$26=u+\dfrac{1}{2}a(2(6)-1)\\\\26=u+5.5a$ ...... ( 1 )

For , n = 7 :

$30=u+\dfrac{1}{2}a(2(7)-1)\\\\30=u+6.5a$ ...... ( 2 )

Solving equation 1 and 2

We get :

$a=4\ m/s^2$ and u = 4 m/s .

Hence , this is the required solution .

Learn More :

Kinematics

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