★ A car is moving with a speed of 10 m/s. When the brakes are applied, the car has a constant acceleration of -5 m/s² . What is the distance travelled before the car comes to rest? Also calculate the time taken by the car to come to rest.
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Answers
using the equation of motion
V = U + at
Your v = 0 m/sec
U = 10 m/sec
t = 20 sec
a in this case negative acceleration i.e. deceleration = ?
so substituong in the equation of motion mentioned above,
Deceleration = 0.5 m/sec^2
Hope you understood
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QUESTION :-
★ A car is moving with a speed of 10 m/s. When the brakes are applied, the car has a constant acceleration of -5 m/s² . What is the distance travelled before the car comes to rest? Also calculate the time taken by the car to come to rest.
GIVEN :-
- Initial velocity , u = 10 m/s
- Final velocity , v = 0 m/s [ Breaks are applied ]
- Acceleration , a = - 5 m/s²
TO FIND :-
- Distance travelled by car , s
- Time taken by car , t
SOLUTION :-
★ using formula of acceleration ★
→ a = v - u/t
- a = acceleration.
- v = final velocity.
- u = initial velocity.
- t = time taken.
→ -5 = 0 - 10/t
→ -5 = -10/t
→ t = -10/-5
→ t = 2 sec.
★ Hence the time taken by the car is 2 seconds.
★ using third equation of motion ★
→ v² - u² = 2as
- v = final velocity.
- u = initial velocity.
- a = acceleration.
- s = distance travelled.
→ (0)² - (10)² = 2 (-5) × s
→ 0 - 100 = -10s
→ -100 = -10s
→ s = -100/-10
→ s = 10
★ Hence the distance travelled by the car is 10 m.