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A train is moving with a velocity of 30m/s.when brakes are applied it is found that the velocity reduces to 10m/s in 240m.when the velocity of train becomes zero,the total distance travelled is?
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Problem: A train is moving with a velocity of 30m/s. When brakes are applied, it is found that the velocity reduces to 10m/s in 240m. When the velocity of the train becomes zero, the total distance travelled is?
Solution:
We can use the equation of motion, v^2 = u^2 + 2as, where v is final velocity, u is initial velocity, a is acceleration, and s is displacement.
Initial velocity, u = 30 m/s
Final velocity, v = 0 m/s
Acceleration, a = (v^2 - u^2)/(2s) = (0 - 30^2)/(2*240) = -3.125 m/s^2 (negative sign indicates deceleration)
To find the total distance travelled, we need to calculate the distance covered during the deceleration phase and the distance covered after the velocity becomes zero.
During the deceleration phase:
Initial velocity, u = 30 m/s
Final velocity, v = 10 m/s
Acceleration, a = (v-u)/t
t = (v-u)/a = (10-30)/(-3.125) = 6.4 s
Displacement, s = ut + 0.5at^2 = 30*6.4 + 0.5*(-3.125)*(6.4)^2 = 96 m
After the velocity becomes zero, the train will continue to move due to its inertia. The distance covered during this phase can be calculated using the equation of motion as follows:
v^2 = u^2 + 2as
0^2 = 30^2 + 2a(s - 240)
s - 240 = -450
s = 690 m
Therefore, the total distance travelled is:
Total distance = 96 + 690 = 786 m
Answer: The total distance travelled by the train is 786 m.
Solution:
We can use the equation of motion, v^2 = u^2 + 2as, where v is final velocity, u is initial velocity, a is acceleration, and s is displacement.
Initial velocity, u = 30 m/s
Final velocity, v = 0 m/s
Acceleration, a = (v^2 - u^2)/(2s) = (0 - 30^2)/(2*240) = -3.125 m/s^2 (negative sign indicates deceleration)
To find the total distance travelled, we need to calculate the distance covered during the deceleration phase and the distance covered after the velocity becomes zero.
During the deceleration phase:
Initial velocity, u = 30 m/s
Final velocity, v = 10 m/s
Acceleration, a = (v-u)/t
t = (v-u)/a = (10-30)/(-3.125) = 6.4 s
Displacement, s = ut + 0.5at^2 = 30*6.4 + 0.5*(-3.125)*(6.4)^2 = 96 m
After the velocity becomes zero, the train will continue to move due to its inertia. The distance covered during this phase can be calculated using the equation of motion as follows:
v^2 = u^2 + 2as
0^2 = 30^2 + 2a(s - 240)
s - 240 = -450
s = 690 m
Therefore, the total distance travelled is:
Total distance = 96 + 690 = 786 m
Answer: The total distance travelled by the train is 786 m.
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A train is moving with a velocity of 30m/s.when brakes are applied it is found that the velocity reduces to 10m/s in 240m.when the velocity of train becomes zero,the total distance travelled is?
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