From the graph. (i) find the velocity of the object at the end of 2.5 seconds. (ii) Calculate the acceleration.

Question

An object is moving along a straight line with uniform acceleration. The following table gives the velocity of the object at various instants of time

Time (in sec.)	0	1	2	3	4	5	6
Velocity (in m/s)	2	4	6	8	10	12	14

Plot the graph. 

From the graph. 

(i) find the velocity of the object at the end of 2.5 seconds. 

(ii) Calculate the acceleration. 

(iii) Calculate the distance covered in the last 4 seconds.

Answer 1

Following is the graph corresponding to the uniformly accelerated motion

a) Velocity at 2.5 seconds is 7 m/s.

b) Acceleration = slope of velocity time graph

= \(\frac{(14-2)}{(6-0)}\)

= 2 m/s²

c) Distance covered in last 4 seconds is the area under the curve for last 4 second 

i.e. 

Total distance covered in last 4 seconds 

= area under ABCD 

= area of ABED + area of BCE

= ((6-0)x(6-2)) +\(\frac{(14-6)\times(6-2)}{2}\)

= 24 + 16

= 40 m