a train accelerated from 20km/h to 80km/h in 4 min. how much distance does it cover in this period. assume that the tracks are straight?
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Answered by
17
Here,
Initial velocity(u) = 20 km/hr
final velocity (v) = 80km/hr
time take (t) = 4/60 hr
= 1/15 hr
Distance(s )= ?
so firstly we have to find acceleration
Acceleration(a) = (v - u)/t
= ( 80-20)/1/15
= 60× 15
= 900 km/hr ^2
Now by using formula
v^2 = u^2 + 2as
(80)^2 = (20)^2 + 2× 900×s
6400 = 400 + 1800 × s
6400-400/1800 = s
6000/1800 = s
60/18 = s
10/3 = s
3.33 = s
s= 3.33 km is your answer
Initial velocity(u) = 20 km/hr
final velocity (v) = 80km/hr
time take (t) = 4/60 hr
= 1/15 hr
Distance(s )= ?
so firstly we have to find acceleration
Acceleration(a) = (v - u)/t
= ( 80-20)/1/15
= 60× 15
= 900 km/hr ^2
Now by using formula
v^2 = u^2 + 2as
(80)^2 = (20)^2 + 2× 900×s
6400 = 400 + 1800 × s
6400-400/1800 = s
6000/1800 = s
60/18 = s
10/3 = s
3.33 = s
s= 3.33 km is your answer
Answered by
6
Explanation:
answer is 10/3km.
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