Question

A car is moving on a straight road with uniform acceleration. The following table gives the speed of the car at various instants of time. ​

Answer 1

Answer:

please follow me and mark me as brainlist

Answer 2

$\huge\red{{SoLuTiOn:—}}$

We take a graph paper and plot the above given time values of 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 the above attachment. Please note 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 the above figure. We will now answer the question.

i) Calculation of acceleration. We know that :

$acceleration = slope \: of \: speed - time \: graph$

$= slope \: of \: line \: af$

$= \frac{fg}{ag}$

Now,if we look at the graph, 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 :

$acceleration = \frac{25m {s}^{ - 1} }{50s}$

$= 0.5 \: {ms}^{ - 1}$

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 . But the figure OAFH is a trapezium. So,

Distance travelled = Area of trapezium OAFH

= (Sum of two parallel sides )×Height /2

In the above figure, the two parallel sides are OA and HF whereas the height is OH . Therefore,

Distance travelled = (OA+HF)×OH / 2

= (5+30)×50 / 2

=35×50 / 2

= 875 m