Question

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 s2. Find
how far the train will
go before it
is brought to rest.

Answer 1

Answer:

Speed of train , u = 90 km/h = 90 × 5/18 = 25 m/s

Speed of train , u = 90 km/h = 90 × 5/18 = 25 m/sacceleration , a = -0.5 m/s²

Speed of train , u = 90 km/h = 90 × 5/18 = 25 m/sacceleration , a = -0.5 m/s²Use formula,

Speed of train , u = 90 km/h = 90 × 5/18 = 25 m/sacceleration , a = -0.5 m/s²Use formula,v = u + at

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 = 0

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 - 0.5t

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

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 secAgain, use formula,

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 secAgain, use formula,S = ut + 1/2at²

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 secAgain, use formula,S = ut + 1/2at²Where S is distance travelled before stop

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 secAgain, use formula,S = ut + 1/2at²Where S is distance travelled before stopS = 25 × 50 - 1/2 × 0.5 × 50²

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 secAgain, use formula,S = ut + 1/2at²Where S is distance travelled before stopS = 25 × 50 - 1/2 × 0.5 × 50²= 1250 - 1/2 × 0.5 × 2500

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 secAgain, use formula,S = ut + 1/2at²Where S is distance travelled before stopS = 25 × 50 - 1/2 × 0.5 × 50²= 1250 - 1/2 × 0.5 × 2500= 1250 - 625

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 secAgain, use formula,S = ut + 1/2at²Where S is distance travelled before stopS = 25 × 50 - 1/2 × 0.5 × 50²= 1250 - 1/2 × 0.5 × 2500= 1250 - 625= 625 m

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 secAgain, use formula,S = ut + 1/2at²Where S is distance travelled before stopS = 25 × 50 - 1/2 × 0.5 × 50²= 1250 - 1/2 × 0.5 × 2500= 1250 - 625= 625 mHence, distance travelled = 625m and time taken = 50 sec

Answer 2

$\huge\star\sf\underline\pink{Solution:-}$

$\sf{ We\:have,} \\ \\ \\ \sf{ Initial\:speed (u) = 90 km/h} \\ \\ \longrightarrow\sf{ \dfrac{90 \times 1000\:m}{60 \times 60\:s} } \\ \\ \longrightarrow\sf{25\:m\:s ^{-1} } \\ \\ \sf{Final \:speed (v) = 0} \\ \\ \sf{Acceleration (a) = -0.5\:m/s^2} \\ \\ \huge\sf\underline\blue{To\:Find:-} \\ \\ \sf{ Distance\:travalled (s) = ? } \: \: \: \sf\purple{(To\:be\: calculated)} \\ \\ \\ \sf\underline\red{Equation\:of\: Motion:-} \\ \\ \underline\pink{\boxed{\sf{ v^2 = u^2 + 2as}}} \\ \\ \sf{So,} \\ \\ \sf{ (0)^2 = (25)^2 + 2 \times (-0.5) \times s} \\ \\ \longrightarrow\sf{ 0 = 625 - 1 \times s } \\ \\ \longrightarrow\underline\orange{\boxed{\sf{s = 625\:m}}} \\ \\ \sf{Thus,} \\ \\ \sf\red{ The\:Train\:will\: travel\:a\: distance\:of\:625\: metres.}$