Question
An express train going towards Delhi is travelling at a speed of 90 kmph. Brakes are applied so as to produce a uniform acceleration of –0.5m s−2 . Find how far the train will go before it is brought to rest.
Answer: Option C
:
C
Given,
initial velocity, u=90kmph = 90×518 =25ms−1
acceleration, a=–0.5ms−2
final velocity, v=0 (brakes are applied)
Let 's' be the distance travelled.
From the third equation of motion,
v2=u2+2as
0=252+(2×−0.5×s)
⇒s=625m
Hence the distance covered is 625m.
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:
C
Given,
initial velocity, u=90kmph = 90×518 =25ms−1
acceleration, a=–0.5ms−2
final velocity, v=0 (brakes are applied)
Let 's' be the distance travelled.
From the third equation of motion,
v2=u2+2as
0=252+(2×−0.5×s)
⇒s=625m
Hence the distance covered is 625m.
Was this answer helpful ?
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