3) A train is travelling at the speed of 90km/h. Brakes are applied so as to produce uniform acceleration of _0.5m/s2. Find how far the train will go before it is brought to rest.
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Answered by
0
Answer:
a train is travelling at the speed 450
Answered by
7
Answer:
Hence, distance travelled = 625m
And time taken = 50 sec
Explanation:
Given,
Initial Speed of train , u = 90 km/h = 90 × 5/18 = 25 m/s
Acceleration , a = -0.5 m/s²
By using first equation of motion formula, we get,
v = u + at
Finally train will be rest so, final velocity, v = 0
0 = 25 - 0.5t
25 = 0.5t ⇒
t = 50 sec
Again, by using third equation of motion formula,
S = ut + 1/2at²
Where S is distance travelled before stop,
S = 25 × 50 - 1/2 × 0.5 × 50²
S = 1250 - 1/2 × 0.5 × 2500
S = 1250 - 625
S = 625 m
Hence, distance travelled = 625m and time taken = 50 sec
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