Question

Following figure shows velocity -time graph of an object. 

a) Sketch the acceleration-time graph that correspond to this motion. 

b) Assuming the initial position is zero, determine the displacement of the object from beginning to t=8.0 seconds after it starts. 

Answer 1

a)

Acceleration in first phase

a1 = 30/2 = 15 m/s^2

Acceleration in second phase

a2 = 0 m/s^

Acceleration in 3rd phase

a3= - 30 /4 = - 7.5 m/s^2

=====

b)

net displacement

d = 0.5* ( 2 + 8)* 30

4 a t -

d = 150 m

======

Comment before rate in case any doubt, will reply for sure.. Goodluck

Answer 2

SOLUTION :

a.

Acceleration :

Between 0 to 2.0 sec , a1 = ∆V / ∆t = 30 /.20       = 15.0 m/s^2 

Between 2.0 to 4.0 sec , a1 = ∆V / ∆t = 0 / 2.0    = 0.00 m/s^2 

Between 4.0 to 8.0 sec , a1 = ∆V / ∆t = - 30 / 4.0 = 7.50 m/s^2 

So the graph of acceleration will be on grid with t on x-axis and acceleration on y-sec with values as under :

          t  (sec)                    acceleration (m/s^2)            

           0 - 2.0                            15.00

           2.0 - 4.0                           0.00

           4.0 - 8.0                        -  7.50 

This graph can easily be drawn.

b. 

Displacement of the object 

= Area of the given graph of velocity v/s time.

= 1/2 * 2.0 * 30 + 2.0 * 30 + 1/2 * 4.0 * 30

= 30 + 60 + 60

= 150 m (ANSWER)