Question

(i) Identify the type of motion by line OA and BC.
(ii) What is the time duration during which the car moves with constant velocity?
(iii) Calculate the acceleration between the 3rd and 10th second.
(v) What is the distance travelled in the first 10 s?​

Answer 1

Answer:

1) 2d motion

2)

Explanation:

5) 4.6 m/s

Answer 2

Answer:

(1) OA represents uniform acceleration since the slope of OA in the velocity-time graph is having a uniform positive slope. AB represents the uniform velocity of 20ms  

−1

. Since the slope of AB is zero. hence, acceleration is zero.

(2) After 10 s, the velocity is 20ms  

−1

 up to 30 s and after 30 s, the velocity is uniformly restarted to zero after 40 s.

(3) Retardation is uniform and it is equal to the slope of BC, i.e.

Slope of BC =  

DC

BD

​

=  

40−30

20

​

=2ms  

−2

 

And acceleration =−2ms  

−2

 

(4) Distance covered by the body between the 10th and 30th second

=20×20=400metres

Explanation: