A train is moving with a velocity of 90 km/h. It is brought to stop by applying brakes which produce a retardation of 0.5 m s-2. Find the velocity after 10 s and time taken by the train to come at rest
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Answered by
124
i) v = 0
a = - 0.5 m/s²
t = ?
u = 90 × 5/18 m/s = 25 m/s
v = u + at
0 = 25 + (- 0.5) t
0= 25 - 0.5t
0.5 t = 25
t = 25 / 0.5
t = 50 s
ii) u = 90 km/h = 90 × 5/18= 25 m/s
a = - 0.5 m /s²
t = 10 s
v = u + at
v = 25 + ( - 0.5) 10
v = 25 - 5 = 20 m/s
a = - 0.5 m/s²
t = ?
u = 90 × 5/18 m/s = 25 m/s
v = u + at
0 = 25 + (- 0.5) t
0= 25 - 0.5t
0.5 t = 25
t = 25 / 0.5
t = 50 s
ii) u = 90 km/h = 90 × 5/18= 25 m/s
a = - 0.5 m /s²
t = 10 s
v = u + at
v = 25 + ( - 0.5) 10
v = 25 - 5 = 20 m/s
Answered by
22
Explanation:
i) v = 0
a = - 0.5 m/s²
t = ?
u = 90 × 5/18 m/s = 25 m/s
v = u + at
0 = 25 + (- 0.5) t
0= 25 - 0.5t
0.5 t = 25
t = 25 / 0.5
t = 50 s
ii) u = 90 km/h = 90 × 5/18= 25 m/s
a = - 0.5 m /s²
t = 10 s
v = u + at
v = 25 + ( - 0.5) 10
v = 25 - 5 = 20 m/s
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