a ball is moving with a velocity of o.5m/s. Its velocity is decreasing at the rate of 0.05m/s. What is the velocity after 5 seconds? how much time will it take from start to stop? What is the distance travelled by it before stopping?
give answe with complete solution please!
Answers
Answer :
- the velocity after 5 seconds = 0.25 m/s
- time it takes from start to stop = 10 sec
- the distance travelled by it before stopping = 2.5 m
Explanation :
Given,
- A ball is moving with a velocity of 0.5 m/s
i.e., initial velocity, u = 0.5 m/s
- Its velocity is decreasing at the rate of 0.05 m/s
i.e., acceleration, a = –0.05 m/s² (since velocity is decreasing)
To find,
- the velocity after 5 seconds
- time it takes from start to stop
- the distance travelled by it before stopping
Solution,
[1] the velocity after 5 seconds :
The first equation of motion is
- v = u + at
where
v - velocity after t sec
u - initial velocity
a - acceleration
Substituting the given values and t = 5 sec,
➙ v = 0.5 + (–0.05)(5)
➙ v = 0.5 - 0.25
➙ v = 0.25 m/s
∴ The velocity after 5 sec = 0.25 m/s
[2] time it takes from start to stop :
Let t sec be the time it takes to stop
final velocity, v = 0
The first equation of motion is
- v = u + at
Substituting the given values,
➙ 0 = 0.5 + (-0.05)(t)
➙ 0 = 0.5 - 0.05t
➙ 0.05t = 0.5
➙ t = 0.5/0.05
➙ t = 10 sec
∴ It will take 10 sec from start to stop.
[3] The distance travelled by it before stopping :
Let 's' be the distance covered by it before stopping
By third equation of motion,
v² - u² = 2as
where
S - distance covered
u - initial velocity
v - final velocity
a - acceleration
Substituting the given values,
➙ 0² - (0.5)² = 2(-0.05)(s)
➙ 0 - 0.25 = -0.1s
➙ -0.25 = -0.1s
➙ 0.1s = 0.25
➙ s = 0.25/0.1
➙ s = 2.5 m
∴ The distance covered by the ball before stopping = 2.5 m
Answer:
the velocity after 5 seconds = 0.25 m/s
time it takes from start to stop = 10 sec
the distance travelled by it before stopping = 2.5 m
Explanation :
Given,
A ball is moving with a velocity of 0.5 m/s
i.e., initial velocity, u = 0.5 m/s
Its velocity is decreasing at the rate of 0.05 m/s
i.e., acceleration, a = –0.05 m/s² (since velocity is decreasing)
To find,
the velocity after 5 seconds
time it takes from start to stop
the distance travelled by it before stopping
Solution,
[1] the velocity after 5 seconds :
The first equation of motion is
v = u + at
where
v - velocity after t sec
u - initial velocity
a - acceleration
Substituting the given values and t = 5 sec,
➙ v = 0.5 + (–0.05)(5)
➙ v = 0.5 - 0.25
➙ v = 0.25 m/s
∴ The velocity after 5 sec = 0.25 m/s
[2] time it takes from start to stop :
Let t sec be the time it takes to stop
final velocity, v = 0
The first equation of motion is
v = u + at
Substituting the given values,
➙ 0 = 0.5 + (-0.05)(t)
➙ 0 = 0.5 - 0.05t
➙ 0.05t = 0.5
➙ t = 0.5/0.05
➙ t = 10 sec
∴ It will take 10 sec from start to stop.
[3] The distance travelled by it before stopping :
Let 's' be the distance covered by it before stopping
By third equation of motion,
v² - u² = 2as
where
S - distance covered
u - initial velocity
v - final velocity
a - acceleration
Substituting the given values,
➙ 0² - (0.5)² = 2(-0.05)(s)
➙ 0 - 0.25 = -0.1s
➙ -0.25 = -0.1s
➙ 0.1s = 0.25
➙ s = 0.25/0.1
➙ s = 2.5 m
∴ The distance covered by the ball before stopping = 2.5 m
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