A particle is pushed along a horizontal surface in such a way that it starts with a velocity of 12 m/s. Its velocity decreases at a uniform rate of 0.5 m/s2 . a) Find the time it will take to come to rest. b) Find the distance covered by it before coming to rest.
Answers
Given :
▪ Initial velocity = 12m/s
▪ Deceleration = 0.5m/s²
To Find :
▪ Time taken by particle to come to rest.
▪ Distance travelled by particle before it is brought to rest.
Concept :
☞ Since, acceleration has said to be constant throughout the motion, we can easily apply equation of kinematics to solve this question.
❇ First equation of kinematics :
❇ Third equation of kinematics :
Calculation :
✴ Calculation of time :
✴ Calculation of distance :
[Note : -ve sign of a shows retardation.]
Given:-
- Initial velocity (u) = 12 m/s .
- Final velocity (v) = 0 m/s
- Acceleration (a) = - 0.5 m/s²
To Find:-
- (a) Time taken to come to rest.
- (b) Distance covered by before coming to rest.
Solution :-
In First case we have to find the time taken to come to rest.
By using 1st equation of motion
☛ v = u + at
putting all the values which is given above
➭ 0 = 12 + (- 0.5 )× t
➭ -12 = (-0.5 (× t
➭ t = 12/0.5
➭ t = 24 s.
∴ The time taken to come to rest is 24 s .
Now, In Second case we have to calculate the distance covered by it before coming to rest.
Using the 2nd equation of motion
☛ s = ut +1/2at²
➭ s = 12 ×24 + 1/2 × (-0.5 )× 24²
➭ s = 288 +1/2 ×(-0.5 )× 576
➭ s = 288 + (-0.5) × 288
➭ s = 288+(-144)
➭ s = 288 - 144
➭ s = 144 m