Question

If a freely falling body travels in the last second a distance equal to the distance travelled by it in the first 3 seconds. Find the time of its travel.
A. $3s$
B. $4s$
C. $5s$
D. $6s$

Answer 1

Apply the kinematic equation of motion, where you substitute the value of $u$ is the initial velocity, $t$ is the time, $g$ is the acceleration of gravity. Force is proportional to mass and acceleration. Calculate the distance travelled for the first 3 seconds. Then calculate the distance travelled at the last second.

Formula:
$s = ut + \dfrac{1}{2}g{t^2}$
$u$ is the initial velocity, $t$ is the time, $g$ is the acceleration of gravity.

Complete step by step answer:
A motion equation helps to describe a body’s location, velocity or acceleration relative to frame of reference.
The velocity and the position can be derived from the newton equation by the method of integration.
Here force acting on a body is known as the function of time. The velocity equation integration results in the distance equation. The velocity equation integration results in the distance equation.
Kinematic equation of motion helps to describe a body’s location, velocity or acceleration relative to frame of reference.
The velocity and the position can be derived from the newton equation by the method of integration.

Here force acting on a body is known as the function of time. The velocity equation integration results in the distance equation.
Distance travelled in first three seconds =$\dfrac{1}{2}g{t^2} = \dfrac{1}{2} \times 9.8 \times 9 = 44.1m$
Time taken to travel the total height is $t$
Then velocity at ${\left( {t - 1} \right)^{th}}$ second = $g \times \left( {t - 1} \right)$
Then we can write distance travelled at the last second= $g\left( {t - 1} \right) - \dfrac{1}{2}g = g\left( {t - 1.5} \right) = 44.1m$

Therefore the time travelled by the body $t = \dfrac{{44.1}}{{9.8}} + 0.5 = 5s$
Hence, the correct answer is option (C).

Note: The initial velocity will be zero if the motion starts from rest and the frame of reference should be the same. The velocity and the position can be derived from the newton equation by the method of integration. The velocity equation integration results in the distance equation.