The velocity time graph of a car of 1000 kg mass is given below: 2 I) When is the acceleration maximum? Why? II) What is the value of retarding force?
Answers
Explanation:
Mass of the car, m = 1000 kg
(a) Using second law :
F = m a
where a is the acceleration of the car, a = v/t. So, the maximum velocity of the car is from 0 to 4 s.
So, F=1000\times \dfrac{15-0}{4}=3750\ NF=1000×
4
15−0
=3750 N
(b) When the acceleration of the car is decreasing then there exits retardation. It is also known as negative acceleration. From the graph the retardation is from 10 s to 12 s.
F=1000\times \dfrac{0-15}{2}=-7500\ NF=1000×
2
0−15
=−7500 N
(c) When the velocity time graph is parallel to x axis then the acceleration of the car is zero. For 6 seconds there is no force acting on the car.
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Answer:
Acceleration is maximum from A to B because slope of OA is max and AB is having constant acceleration .
Also a=15-0/4-0=15/4 m/s^2.
Value of retarding force=mass×retardation
Force=1000×(0-15/12-10)=1000×-15/2
= -7500 N.