Question

A body is thrown vertically upwards with $55m/s$ What is the distance travelled in the $6th$ second?

Answer 1

We start this question by finding the displacement travelled in five seconds and the distance travelled in six seconds. We subtract these values. Then we find the time at which the body reaches its peak magnitude. By adding the distance travelled by the body in five seconds to the peak time and the peak time to six seconds.

Formulas used: The equation of motion used to find the displacement is given by $S = ut + \dfrac{1}{2}a{t^2}$
The formula to find the acceleration is given by $a = \dfrac{{v - u}}{t}$
Where $v$ is the final velocity
$u$ is the initial velocity
$t$ is the time taken

Complete step by step answer:
Let us start by noting down the given data
The initial velocity is given as $u = 55m/s$
The acceleration of the body is $a = - 9.8$
We can find the time at which the body is at its peak position, that is when the final velocity is zero
we can use the formula of acceleration and find the time by $t = \dfrac{{v - u}}{a}$
substituting the values, we get $t = \dfrac{{v - u}}{a} = \dfrac{{0 - 55}}{{ - 9.8}} = 5.61s$
We know that we have the time at peak value, we find the distance travelled by the body in this time and five and six seconds.
For five seconds $S = ut + \dfrac{1}{2}a{t^2} = 55 \times 5 + \dfrac{1}{2} \times - 9.8 \times {5^2} = 152.5m$
For the peak value $S = ut + \dfrac{1}{2}a{t^2} = 55 \times 5.61 + \dfrac{1}{2} \times - 9.8 \times {5.61^2} = 154.34m$
For six seconds $S = ut + \dfrac{1}{2}a{t^2} = 55 \times 6 + \dfrac{1}{2} \times - 9.8 \times {6^2} = 153.6m$
The distance travelled by the sixth second is asked, so we subtract the distance travelled by peak value from the distance travelled in six seconds and add this with the difference of the displacement.
Distance travelled by the body from the fifth to the peak displacement time is $154.34 - 152.5 = 1.84$
Distance travelled by the body from the peak displacement time and sixth second $154.34 - 153.6 = 0.74$
The total distance in the sixth second is, $0.74 + 1.84 = 2.58m$

Note: We are asked to find the distance travelled in the sixth second and not the displacement. If we were asked to find the displacement, then we could have directly applied the formula and got the value. Distance means the total amount of length travelled by the body and displacement is the difference between the initial and final position.