A train is travelling at a speed of 90 km h-1 . Brakes are applied so as to produces a uniform acceleration of -0.5 ms-2 .find how far the train will go before it is brought to rest.
Answers
Given :-
Initial velocity of train = 90 Km/h = 25 ms-¹
De-acceleration = a = - 0.5 ms-²
When a body is travelling then it has initial velocity only.
Using third equation of Motion :
v² - u² = 2aS
-(25)² = 2(-0.5)S
S = 625 m
Alternative way,
A formula is used when brakes are applied,
S = u²/2a
Here we can neglect negative sign in Acceleration.
S = (25)²/2(0.5)
S = 625/1
S = 625 m.
Hence,
Distance travelled = S = 625 m
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