Question

41. A body, moving in a straight line, with an
initial velocity u and a constant acceleration a,
covers a distance of 40 m in the 4th second and
- a distance of 60 m in the 6th second. The values
of u and a respectively are
(a) 10 m s-1,5 m s-2 (b) 10 m s-1, 10 m s-2
(c) 5 m s -1,5 m s-2 (d). 5 m s-1, 10 m s-2​

Answer 1

correct option is d . here is ur answer

Answer 2

Given:

• Displacement in 4th sec = 40 m
• Displacement in 6th sec = 60 m

To find:

• Value of initial velocity and acceleration?

Calculation:

For nth second, the displacement is:

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

At 4th sec , we can say :

$\implies 40= u + \dfrac{1}{2} a(2 \times 4 -1 )$

$\implies 40 = u + \dfrac{7a}{2}$

At 6th second:

$\implies 60= u + \dfrac{1}{2} a(2 \times 6-1 )$

$\implies 60 = u + \dfrac{11a}{2}$

Now, subtraction of the equations:

$60 - 40 = \dfrac{11a - 7a}{2}$

$\implies 20 = \dfrac{4a}{2}$

$\implies a = 10 \: m {s}^{ - 2}$

So, acceleration is 10 m/s².

Now, putting value of a :

$\implies 60 = u + \dfrac{11(10)}{2}$

$\implies 60 = u + 55$

$\implies u = 5 \: m {s}^{ - 1}$

So, initial velocity is 5 m/s

OPTION D) IS CORRECT ✔️