# A person travelling at 43.2 km/h applies the brake giving a decelerations of 6.0 m/s2 to his scooter. How far will it travel before stopping?

Navjyot Kalra
8 years ago
Sol. Initial velocity u = 43.2 km/hr = 12 m/s u = 12 m/s, v = 0 a = –6 m/s2 (deceleration) Distance S = (V^2-u^2)/(2 (-6)) = 12 m
Parth
22 Points
8 years ago
the magnitude of two vector are equal and the angle between them is θ. show that their resultant divides angle θ equally
Parth
22 Points
8 years ago
the magnitude of two vector are equal and the angle between them is θ. show that their resultant divides angle θ equally
A
13 Points
3 years ago
The correct answer is 6m.                                                                                          1km/hr=18/5 m/s so 43.2km/hr=12m/s
u=12m/s v=0m/s a=negative 12m/s^2(as it is a retardation)
Now, v=u+at substitute the values then t=1s
Finally, using the formula
s=ut+1/2at^2 substitute the values to get s=(12)(1)+1/2(-12)(1)^2 so the answer is 6m
Razz kumar
15 Points
3 years ago
Now
V=0m/s
u=43.2m/s
a=(-6)m/s²
So now
according to formula
V²=u²+2as.
0²=(43.2×5/18)²+2×(-6)×s.
0=144-12s.
-144=-12s
S=12m