Question

A body falls for 4 seconds from rest. If the acceleration due to gravity ceases to act, then the velocity and distance it travels in the next 3sec

Answer 1

hello friend

lly the body starts its motion from rest therefore u = 0
let this journey be A, when accelaration due to earth is acting on the body
time = 5seconds. Let s find v
v= u+ at
= 0 + 9.8 x 5
=49m/s

when accelaration stops acting the velocity of the body is 49m/s
now the distance travelled by the body in the next 3s is

s= ut + 1/2gt2
= 49 x 3 + ½ x 0 x t2
= 49 x 3

147m
good bye

Answer 2

The velocity and the distance is 40 m/s and 120 m respectively

A body is dropped from some height and it falls for a certain distance for a time period of 4 s and then once the acceleration due to gravity (g) ceases it continues to fall for the next 3 s.

The body is said to fall from rest which means that the body was initially at rest with no initial velocity. The initial velocity (u) of the body initially is hence zero.

⇒$u=0$

The body falls with some velocity which is variable in nature and hence keeps changing. This variable velocity is due to the acceleration due to gravity that was present for the first 4 seconds.

Thus,  $u=0\;m/s$  and $t=4s$

Taking acceleration due to gravity as $a=10\;m/s^{2}$ as it a constant value

The equation for kinematics is given below:

$v=u+a \times t$

where,

v is the velocity of the body

u is the initial velocity of the body

t is the time period

a is the acceleration due to gravity

By substituting the values of initial velocity, time period and acceleration due to gravity in the equation we get:

$v=0+(10)(4)$

$\implies v=40$

Hence, we get the velocity as $v=40\;m/s$

It is said that the acceleration due to gravity ceases after a time period of 4 seconds which means that the acceleration due to gravity is not present anymore.

This results in the body traveling with a constant velocity now. Since there is no acceleration due to gravity in the next time period of 3 seconds, the velocity is constant enabling the body to fall with a stable speed.

We are asked to calculate the distance for the next 3 seconds after acceleration due to gravity ceases.

We use the formula given below to compute the distance:

$Distance=Speed \times Time$

W.K.T

$v=40$ and $t=3s$

By substitution of the velocity and time period values we get:

$Distance=40 \times 3$

$\implies Distance=120\;m$

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