A car moving with a velocity of 50 Km/hr can be stopped by breaks after at least 6 m. If the same car is moving
at a speed of 100 Km/hr, the minimum stopping distance is

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Solution

Initial velocity, u = 50 km/h = 50 Ã— (5/18) = 250/18 m/s
Final velocity, v = 0
Distance travelled before coming to rest, s = 6 m
Using, v² = u² + 2as
=> a = -u²/(2s)
=> a = -16.075 m/s²
Again,
u = 100 km/h = 500/18 m/s
v = 0
a = -16.075 m/s²
Now, v² = u² + 2as
=> s = -u²/(2a) = 24 m

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A car moving with a velocity of 50 Km/hr can be stopped by breaks after at least 6 m. If the same car is movingat a speed of 100 Km/hr, the minimum stopping distance is

Initial velocity, u = 50 km/h = 50 Ã— (5/18) = 250/18 m/sFinal velocity, v = 0Distance travelled before coming to rest, s = 6 mUsing, v2 = u2 + 2as=> a = -u2/(2s)=> a = -16.075 m/s2Again,u = 100 km/h = 500/18 m/sv = 0a = -16.075 m/s2Now, v2 = u2 + 2as=> s = -u2/(2a) = 24 m

A car moving with a velocity of 50 Km/hr can be stopped by breaks after at least 6 m. If the same car is moving
at a speed of 100 Km/hr, the minimum stopping distance is

Initial velocity, u = 50 km/h = 50 Ã— (5/18) = 250/18 m/s
Final velocity, v = 0
Distance travelled before coming to rest, s = 6 m
Using, v² = u² + 2as
=> a = -u²/(2s)
=> a = -16.075 m/s²
Again,
u = 100 km/h = 500/18 m/s
v = 0
a = -16.075 m/s²
Now, v² = u² + 2as
=> s = -u²/(2a) = 24 m