A train comes to stop in 90 m at an acceleration of 3 m/s2 . Find the speed at which it was travelling About the author Josephine
Answer: Given that, Acceleration a=−0.5m/s2 Speed v=90km/h=25m/s Using equation of motion, v=u+at Where, v = final velocity u = initial velocity a = acceleration t = time Put the value into the equation Finally train will be rest so, final velocity,v=0 0=25−0.5t 25=0.5t t=0.525 t=50 sec Again, using equation of motion, S=ut+21at2 Where, s = distance v = final velocity u = initial velocity a = acceleration t = time Put the value into the equation Where S is distance travelled before stop s=25×50−21×0.5×(50)2 s=625 m So, the train will go before it is brought to rest is 625 m. Hence, A is correct. Reply
Answer:
Given that,
Acceleration a=−0.5m/s2
Speed v=90km/h=25m/s
Using equation of motion,
v=u+at
Where,
v = final velocity
u = initial velocity
a = acceleration
t = time
Put the value into the equation
Finally train will be rest so, final velocity,v=0
0=25−0.5t
25=0.5t
t=0.525
t=50 sec
Again, using equation of motion,
S=ut+21at2
Where, s = distance
v = final velocity
u = initial velocity
a = acceleration
t = time
Put the value into the equation
Where S is distance travelled before stop
s=25×50−21×0.5×(50)2
s=625 m
So, the train will go before it is brought to rest is 625 m.
Hence, A is correct.