Q1.A car starts from rest and attains a velocity of 10 metre per second in 40
seconds .the driver applies brakes and slows down the car to 5 metre per
second in 10 seconds. Find the acceleration of the car in both the cases?
Answers
Explanation:
✬ Acceleration = 0.25 & 0.5 m/s² ✬
Explanation:
Given: (Solving first case)
A car starts from rest.
Final velocity of car is 10 m/s.
Time taken to attain final velocity is 40 s.
To Find:
Acceleration of the car ?
Formula to be used:
V = u + at
Solution: Here we have
u = 0
v = 10 m/s
t = 40 s
a = ?
Substituting the values on formula
⟹ 10 = 0 + a × 40
⟹ 10 = 40a
⟹ 10/40 = a
⟹ 1/4 = a
⟹ 0.25 m/s² = a
So acceleration of car in first case is 0.25 m/s².
Given: (Second case)
Final velocity is 5 m/s.
Initial velocity is 10 m/s.
Time given is 10 s.
To Find:
Acceleration of the car ?
Formula to be used:
a = v – u/t
Solution: Here we have
u = 10 m/s
v = 5 m/s
t = 10 s
a = ?
Substituting the values on formula
⟹ a = 5 – 10/10
⟹ a = –5/10
⟹ a = –1/2
⟹ a = –0.5 m/s²
So, acceleration of car in second case is –0.5 m/s².
(You can also solve this one by using Newton's first law of motion , used in above case)
___________________
- The velocity at which motion starts is termed as initial velocity. It is denoted by u.
- The last velocity of an object after a period of time. It is denoted by v.
- v = ut + 1/2at² ( Second law of motion)
- v² = u² + 2as ( Third law of motion )
Answer:
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