A bmw car travelling at 22.4 litre per second squared to stop in 2.55 seconds determine the skidding distance of the car assume uniform acceleration
Answers
HEYA GM ❤
★ STOPPING DISTANCE = U² / 2 a
a = Acceleration , U = speed
- From first EQUATION Of MOTION ...
V = U + at
V = 22.4 × 2.55 = 57.12 m/s ..
★ Stopping Distance =( 57.12)² / 2 (22.4)
= 3262.7 / 44.8
= 72.828 m
» Car Stopped after traveling 72.8 Metres...
TQ ❤ ❤
Answer:
Explanation:
Known Terms :-
u = Initial speed
v = Final speed
a = Acceleration
S = Distance
t = Time
Given :-
Initial speed of the car, u = 22.4 m/s
Final speed of the car, v =0 m/s (Because car skids and stops)
Time taken by the car, t = 2.55 sec
To Calculate :-
Skidding Distance of the car (s) = ??
Formula to be used :-
v = u + at and v² = u² + 2as
Solution :-
1st we have to find the acceleration
Putting all the values , we get
⇒ v = u + at
⇒ 0 = 22.4 + 2.55 × a
⇒ -2.55a = 22.4
⇒ a = 22.4/2.55
⇒ a = -8.784 m/s²
Now we have acceleration
So we will put value in distance formula.
⇒ v² = u² + 2as
⇒ 0 = (22.4)² + 2 × (-8.784) × s
⇒ -501.76 = -17.568 × s
⇒ s = 28.56 meters
Hence, car takes 28.56 meters before its stops.