a train travelling at a speed of 90 km / h . Brakes are applied acceleration of 0.5 m/ s2 ? find how far the train will go before it is brought to rest ?
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4
Heya user !!!
Here's the answer you are looking for
Initial velocity of the train (u) = 90km/he
= 25m/s
Acceleration (a) = -0.5m/s² (negative because it opposes the motion, retardation)
Final velocity (v) = 0 (as it comes to rest)
Using the 3rd equation of motion,
v² - u² = 2as
0² - 25² = 2(-0.5)(s)
-625 = -s
s = 625m
Therefore, the train will travel 625m before coming to rest.
★★ HOPE THAT HELPS ☺️ ★★
Here's the answer you are looking for
Initial velocity of the train (u) = 90km/he
= 25m/s
Acceleration (a) = -0.5m/s² (negative because it opposes the motion, retardation)
Final velocity (v) = 0 (as it comes to rest)
Using the 3rd equation of motion,
v² - u² = 2as
0² - 25² = 2(-0.5)(s)
-625 = -s
s = 625m
Therefore, the train will travel 625m before coming to rest.
★★ HOPE THAT HELPS ☺️ ★★
Answered by
4
90 kmph = 90 × 5/18 m/s = 25 m/s
S = (v² - u²) / (2a)
= (0 - 25²) / (2 × -0.5 m/s²)
= 625 m
Train will cover a distance of 625 m before it comes to rest.
[Note: I took a = -0.5 m/s² because breaks are applying acceleration in the direction opposite to then motion of train].
S = (v² - u²) / (2a)
= (0 - 25²) / (2 × -0.5 m/s²)
= 625 m
Train will cover a distance of 625 m before it comes to rest.
[Note: I took a = -0.5 m/s² because breaks are applying acceleration in the direction opposite to then motion of train].
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