Question

A body is thrown vertically upwards with `55` `ms^-1`. What is the distance travelled in the `6th` second?

Answer 1

Finding the difference between the position at `t=5` and `t=6` `s` we discover that the displacement is `1.1` `m` but the distance traveled is `2.59` `m`. This is because the peak of the trajectory occurs during the `6th` second.

Explanation:

At `t=5` `s`:

`s=ut+1/2at^2 = 55(5)+1/2(-9.8)5^2`
`=275-122.5=152.5` `m`

At `t=6` `s`:

`s=ut+1/2at^2 = 55(6)+1/2(-9.8)6^2`
`=330-176.4=153.6` `m`

Subtracting the distance at 5 from the distance at 6, we see:

`153.6-152.5=1.1` `m`

This might seem surprising, given the initial velocity, but the object is approaching the peak of its trajectory.

Just for interest, let's calculate the time at the peak.

At the highest point, the velocity is zero.

Use the equation:

`v=u+at`

Rearrange to make `t` the subject:

`t=(v-u)/a=(0-55)/(-9.8)=5.61` `s`

Wait: that means that the object peaked during the 6th second! It's a bit a of a trick question, and the method outlined above is not accurate. The object actually moved upward somewhat and downward somewhat during the 6th second, so the total distance traveled is greater than we calculated.

We need a more complicated approach. Let's find the height at the peak, when `t=5.61` `s`:

`s=ut+1/2at^2=55(5.61)+1/2(-9.8)(5.61)^2=308.55-154.21=154.34` `m`

So from `t=5` to `t=5.61` the object moves up from `152.5` to `154.34`, a distance of `1.85` `m`.

From `t=5.61` to `t=6`, it moves downward from `154.34 to 153.6`, a distance of `0.74` `m`.

Adding these distances together, we get `2.59` `m`.

Why the different answers? The question asks us the distance traveled in the `6th` second, which is `2.59` `m`. If we were asked the displacement: the difference between its positions at `t=5` and `t=6`, the correct answer would be `1.1` `m`.