Question

The initial velocity of a body moving along a straight line in 7 m/s. It has a uniform acceleration of 4 m/s2. The distance covered by the body in the 5th second of its motion is

Answer 1

the distance covered by the body in 5th second = u+a ( n-1/2)= 7+ 4 (5-1/2)
= 7+4×9/2
= 7+ 18 = 25 m

Answer 2

Answer: The distance covered by the body is 85 m.

Explanation:

Given that,

Acceleration $a = 4 m/s^{2}$

Initial velocity u = 7 m/s

Time t = 5 second

Using equation of motion

$s = ut +\dfrac{1}{2}at^2$......(I)

Put the value in equation (I)

The distance in 5 sec

$s_{5} = 7\times 5+\dfrac{1}{2}\times 4\times 25$

$s_{5} = 85\ m$

The distance in 4 sec

$s_{4} = 7\times 4+\dfrac{1}{2}\times 4\times 16$

$s_{4} = 60\ m$

Now, the distance covered by the body in 5th sec

$s = s_{5}-s_{4}$

$s = 85\ m-60\ m$

$s = 25\ m$

Hence, The distance covered by the body is 85 m.