A body starts over a horizontal surface with an intial velocity of 0.5 m/s. Due to friction, its velocity decreases at the rate of 0.05 m/s² (acceleration, -0.05 m/s ). How much time will it take for the body to stop ?
Answers
Given:-
→ Initial velocity of the body = 0.5 m/s
→ Acceleration of the body = -0.05 m/s²
To find:-
→ Time taken by the body to stop.
Solution:-
In this case :-
• Since the body finally stops, so it's final velocity (v) will be zero.
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Now, as per the 1st equation of motion :-
v = u + at
Where :-
• v is the final velocity of the body.
• u is initial velocity of the body.
• a is acceleration of the body.
• t is time taken.
Substituting values, we get :-
⇒ 0 = 0.5 + (-0.05)t
⇒ 0 - 0.5 = -0.05t
⇒ -0.5 = -0.05t
⇒ t = -0.5/-0.05
⇒ t = 10s
Thus, the body takes 10 seconds to stop.
Answer:
Given :-
- A body starts over a horizontal surface with an initial velocity of 0.5 m/s.
- Due to friction it's velocity decreases at the rate of 0.05 m/s² (acceleration is - 0.05 m/s).
To Find :-
- What is the time will it take for the body to stop.
Formula Used :-
As we know that,
➲ v = u + at
where,
- v = Final Velocity
- u = Initial Velocity
- a = Acceleration
- t = Time
Solution :-
Given :
- Final Velocity (v) = 0 m/s
- Initial Velocity (u) = 0.5 m/s
- Acceleration (a) = - 0.05 m/s²
According to the question by using the formula we get,
↦ 0 = (0.5) + (- 0.05) × t
↦ 0 = 0.5 - 0.05 × t
↦ 0 = 0.5 - 0.05t
↦ 0 + 0.05t = 0.5
↦ 0.05t = 0.5
↦ t = 0.5/0.05
↦ t = 5 × 100/5 × 10
↦ t = 500/50
↦ t = 50/5
➠ t = 10 seconds
∴ The body takes 10 seconds to stop the body.
EXTRA INFORMATION :-
★ First Equation Of Motion :
- v = u + at
★ Second Equation Of Motion :
- v² = u² + 2as
★ Third Equation Of Motion :
- s = ut + ½at²