Ques, A Car is moving in a straight the road with uniform acceleration. The following table her the good of the
car at various tables gives the speed of the car at
various instance of this
0-5,20-10,30-15,40-20,50-30
Find the: (i) Accelogestion,
(ii) Distance in 50 seconds
plzz answer fast
I will mark as brainliest answer the correct answer.
Answers
Answer:
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. <br> (i) Calculation of Acceleration. We know that : <br> Acceleration = Slope of speed-time graph <br>
= Slope of line AF (see Figure 42) <br>
<br> 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. <br> Therefore,
FG= 30-5 <br> = 25 m/s <br> Again, at point G, the value of time is 50 seconds whereas that at point A is 0 second. <br> Thus,
AG= 50-0 <br>
= 50 s <br> Now, putting these values of FG and AG in the above relation, we get : <br>
Acceleration =
<br>
<br> (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, <br> Distance travelled = Area of trapezium OAFH <br>
<br> In Figure 42, the two parallel sides are OA and HF whereas the heights is OH. Therefore, <br> Distance travelled =
<br>
<br>
ltBrgt
= 875 m