Question

A body starting from rest has an acceleration of 5m/s2.calculate the distance travelled by it in 4th second

Answer 1

Answer:

$\large{\underline{\boxed{\sf Distance = 17.5\: m}}}$

Explanation:

Given that ;

Initial velocity (u) = 0 m/s

Acceleration (a) = 5 m/s²

To find:

• The distance travelled in the 4th seconds.

Solution:

We know that, the distance travelled(S) in nth second is given by -

Sₙ = u + $\sf \dfrac{a}{2}$ * (2n - 1)

⇒ S₄ = 0 + $\sf \dfrac{5}{2}$ * (2 * 4 - 1)

⇒ S₄ = $\sf \dfrac{5}{2}$ * 7

⇒ S₄ = $\sf \dfrac{35}{2}$

⇒ S₄ = 17.5 m

_______________________

Hence, the distance travelled in the 4th second is 17.5 m.

Answer 2

ANSWER:

distance travelled by the body in 4th seconds = 17.5 meter

Explanation:

given that,

The body starting from rest has an acceleration of 5m/s²

here,

initial velocity of the body = 0 m/s

because it started from rest

and

acceleration of the body during motion = 5 m/s²

now,

given time for which he have to cover distance = 4 seconds

now,

we have,

initial velocity(u) = 0 m/s

acceleration = 5 m/s²

let the distance travelled in 4th seconds be s

so,

time = 4 s

now,

we know that,

distance travelled in nth second

s = u + a/2(2n - 1)

so,

here n = 4

putting the values,

s = 0 + 5/2(2(4) - 1)

s = 5/2(8 - 1)

s = 5/2 × 7

s = 35/2

s = 17.5 m

so,

__________________

distance travelled by the body in 4th seconds

= 17.5 meter

__________________

EXTRA INFORMATION

➡ When any object or body starts it's motion from rest and time is given

then distance travelled = at²

where,

a = acceleration

t = time

➡ When the final velocity is lesser than the initial velocity then acceleration will always negative.

this type of acceleration is also called Retardation.