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A train is travelling at a speed of 90 km h−1. Brakes are applied so as to produce a uniform acceleration of −0.5 m s−2. Find how far the train will go before it is brought to rest.

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Speed of train , u = 90 km/h = 90 × 5/18 = 25 m/sacceleration , a = -0.5 m/s²Use formula,v = u + atFinally train will be rest so, final velocity, v = 00 = 25 - 0.5t25 = 0.5t ⇒t = 50 sec Again, use formula,S = ut + 1/2at²Where S is distance travelled before stopS = 25 × 50...
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Speed of train , u = 90 km/h = 90 × 5/18 = 25 m/s
acceleration , a = -0.5 m/s²
Use formula,
v = u + at
Finally train will be rest so, final velocity, v = 0
0 = 25 - 0.5t
25 = 0.5t ⇒t = 50 sec

Again, use formula,
S = ut + 1/2at²
Where S is distance travelled before stop
S = 25 × 50 - 1/2 × 0.5 × 50²
= 1250 - 1/2 × 0.5 × 2500
= 1250 - 625
= 625 m

Hence, distance travelled = 625m and time taken 50 sec

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Completed graduation in Mathematics from Calcutta University

625m
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Here , u = 90 km = 90× 5/18 m/s= 25 m/s, a= -0.5 , S = ?S = 625 m
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Tutor

U= (90x5/18)=25m/s V= 0 a= -0.5 unit V(square) = U(square) + 2aS 0=25x25 + 2x(-0.5)x S S=625m
Comments

Given: Speed of train , u = 90 km/h = 90 × 5/18 = 25m/sacceleration , a = -0.5m/s²Therefore,v = u + atFinally train will be rest so, final velocity, v = 00 = 25 - 0.5t25 = 0.5t ⇒t = 50sAgain, use formula,S = ut + 1/2at²Where S is distance travelled before stopS = 25 × 50...
read more
Given:
Speed of train , u = 90 km/h = 90 × 5/18 = 25m/s
acceleration , a = -0.5m/s²
Therefore,
v = u + at
Finally train will be rest so, final velocity, v = 0
0 = 25 - 0.5t
25 = 0.5t ⇒t = 50s

Again, use formula,
S = ut + 1/2at²
Where S is distance travelled before stop
S = 25 × 50 - 1/2 × 0.5 × 50²
= 1250 - 1/2 × 0.5 × 2500
= 1250 - 625
= 625m

Hence, distance travelled = 625m and time taken = 50s
read less
Comments

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