A train starting from stationary position and moving with uniform accelaration attains a speed of 36km/hour in 10 minutes.what is the distance travelled?
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initial velocity (u) = 0 m/s
final velocity (v) = 36 km/h (10 m/s)
time (t) = 10 min (600 sec)
acceleration (a) = (v-u)/t
= (10-0)/600
= 10/600
= 0.16 m/s^2.
therefore , distance,
2as = v^2-u^2
2(0.16)(s) = (10)^2-(0)^2
0.32 s = 100-0
0.32 s=100
s=100/0.32
s=312.5m/s^2.
final velocity (v) = 36 km/h (10 m/s)
time (t) = 10 min (600 sec)
acceleration (a) = (v-u)/t
= (10-0)/600
= 10/600
= 0.16 m/s^2.
therefore , distance,
2as = v^2-u^2
2(0.16)(s) = (10)^2-(0)^2
0.32 s = 100-0
0.32 s=100
s=100/0.32
s=312.5m/s^2.
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