Question

What is the magnitude of the average acceleration of a skier who, starting from rest, reaches a speed of $8\text{ m/s}$ when going down a slope for 5 seconds and how far does the skier travel in this time?

Answer 1

We are going to use the concept and formulas of ${{v}_{t}}={{v}_{0}}+{{a}_{t}}t$ and ${{s}_{t}}={{v}_{0}}t+\dfrac{1}{2}{{a}_{t}}{{t}^{2}}$ to find the relation between velocity and acceleration. We use the starting velocity as 0. Putting the values, we get the solutions of the problems.

Complete step-by-step solution:
We are going to use the concept of distance covered, velocity and the acceleration. We denote these attributes as $s,v,a$ respectively.
The skier, starting from rest, reaches a speed of $8\text{ m/s}$ when going down a slope for 5 seconds. His velocity and distance covered at time t is denoted by ${{v}_{t}},{{s}_{t}}$ respectively.
He started from rest which means his initial velocity was 0. We denote it as ${{v}_{0}}=0$.
At 5 seconds his speed becomes $8\text{ m/s}$ and we assume the acceleration was $a$.
Therefore, ${{v}_{5}}=8$.
We know the relation between acceleration and velocity as ${{v}_{t}}={{v}_{0}}+{{a}_{t}}t$.
For $t=5$, we get ${{v}_{t}}={{a}_{t}}t$ which gives ${{a}_{t}}=\dfrac{{{v}_{t}}}{t}$.
We now put the value of $t=5$ in ${{a}_{t}}=\dfrac{{{v}_{t}}}{t}$ to get the acceleration.
So, ${{a}_{5}}=\dfrac{{{v}_{5}}}{5}=\dfrac{8}{5}=1.6$.
Now we find the distance the skier has already travelled.
We know the relation among distance, acceleration and velocity as ${{s}_{t}}={{v}_{0}}t+\dfrac{1}{2}{{a}_{t}}{{t}^{2}}$.
Putting the values, we get ${{s}_{5}}=\dfrac{1}{2}\times 1.6\times {{5}^{2}}=20$ metres.
He has already travelled 20 metres.

Note: We can also use the constant value of the acceleration where $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=a$. The above-mentioned formulas are the derivation of the main function. The integration also gives the consecutive values of velocity and then distance covered.