A train is travelling at 90km/h. Breaks are applied so as to produce a uniform acceleration of - 0.5 m/s sq. Find how far will the train go before it is brought to rest.
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By the application of Third equation of Motion:-
v²-u²= 2aS
Convert 90km/h into m/s
(90x5/18)m/s= 25m/s
v²-u²=2aS
v= Final velocity
u= initial velocity
a= acceleration
S= distance
(25²-0²)m/s= (2x0.5)m/s² x S
(625 -0)m/s= 1 x S
625/1 = S
or, S = 625/1
or, S = 625
⭐Be Brainly⭐
v²-u²= 2aS
Convert 90km/h into m/s
(90x5/18)m/s= 25m/s
v²-u²=2aS
v= Final velocity
u= initial velocity
a= acceleration
S= distance
(25²-0²)m/s= (2x0.5)m/s² x S
(625 -0)m/s= 1 x S
625/1 = S
or, S = 625/1
or, S = 625
⭐Be Brainly⭐
aaravshrivastwa:
Yeha
Your answer is wrong
dont be so happy i was just joking!!!
See it carefully
Answer is 625 metres
What is the answer
See, you have multiplied 2 two times on right side
Haha
Now you mouth is closed .
Okay I will re-edit
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