A car is moving on a striaght road with uniform acceleration. The following table gives the speed of the car at various instants of time :
`{:(,"Speed"(m//s)" ":,,5,,10,,15,,20,,25,,30),(,"Time"(s)" ":,,0,,10,,20,,30,,40,,50):}`
Draw the speed-time graph by choosing a convenient scale. Determine from it :
(i) the acceleration of the car.
(ii) the distance travelled by the car in 50 seconds.

Explanation

We take a graph paper and plot the above given time values on the x-axis. The corresponding speed values are plotted on the y-axis. The speed-time graph obtained from the given readings is shown in Figure 42. Please not that in this case, when the time is 0, then the speed is not 0. The body has an initial speed of 5m/s which is represented by point A in Figure 42. We will now answer the questions asked in this sample problem.
(i) Calculation of Acceleration. We know that :
Acceleration = Slope of speed-time graph
`" "` = Slope of line AF (see Figure 42)
`" "=(FG)/(AG)`
Now, if we look at the graph shown in Figure 42, we will find that the value of speed at point F is 30 m/s and that at point G is 5 m/s.
Therefore, `" "` FG= 30-5
= 25 m/s
Again, at point G, the value of time is 50 seconds whereas that at point A is 0 second.
Thus, `" "` AG= 50-0
`" "` = 50 s
Now, putting these values of FG and AG in the above relation, we get :
`" "` Acceleration = `(25 m//s)/(50s)`
`" "= 0.5 m//s^(2)`
(ii) Calculation of Distance Travelled. The distance travelled by the car in 50 seconds is equal to the area under the speed-time curve AF. That is, the distance travelled is equal to the area of the figure OAFH (see Figure 42). But the figure OAFH is a trapezium. So,
Distance travelled = Area of trapezium OAFH
`" "=(("Sum of two parallel sides")xx"Height")/(2)`
In Figure 42, the two parallel sides are OA and HF whereas the heights is OH. Therefore,
Distance travelled = `((OA+HF)xxOH)/(2)`
`" "=((5+30)xx50)/(2)`
`" "=(35xx50)/(2)` ltBrgt `" "`= 875 m