Question

A body that starts from rest accelerates uniformly along a straight line at the rate of $$10m{s}^{-2}$$ for $$5\sec$$ . It moves for $$2\sec$$ with a uniform velocity of $$50m{s}^{-1}$$ . Then it retards uniformly and comes to rest in $$3\sec$$ . Draw the velocity-time graph of the body and find the total distance travelled by the body.

Answer 1

Hint: The distance travelled is shown by the area under the speed-time graph. For the motion of a particle, distance-time graphs and acceleration-time graphs can be produced, with time always displayed on the horizontal axis. A particle with a constant speed is represented by a straight line on a distance-time graph.

Complete step by step answer:
Let us first write the given values
Acceleration $\left(a\right)=10m{s}^{-2}$
Time $\left(t\right)$ is given as $5\sec$
Therefore the velocity can be found by using equation of motion that is given by:
$$\begin{gathered} v=u+at \\ \Rightarrow v=0+10\times 5 \\ \Rightarrow v=50m{s}^{-1} \end{gathered}$$
Now, it moves for $$2$$ seconds with a uniform velocity of $$50m{s}^{-1}$$ and finally gets retarded to rest in $3\sec$.Using the above information we will draw velocity time graph accordingly;

Now, according to the graph: the total distance travelled is $$\left(D\right)$$ = Area under the curve. Therefore,
$D={\displaystyle \frac{1}{2}}\times \left(OC+AB\right)\times AD$
$$\begin{gathered} \Rightarrow D={\displaystyle \frac{1}{2}}\times \left(10+2\right)\times 50 \\ \Rightarrow D={\displaystyle \frac{1}{2}}\times 12\times 50 \\ \therefore D=300m \end{gathered}$$
Hence, the total distance travelled by the body is $300m$.

Note: It's important to remember that finding the area doesn't fully describe displacement because, being a vector quantity, displacement also requires a direction.Finding the location simply provides a number, not a direction. The magnitude of the displacement, which is equivalent to the distance travelled, is represented by the area under the curve (only for constant acceleration).