Question

A train starting from stationary position attains aspeed of 36 km/h in 10 minute.find its accele4ation

Answer 1

here is your answer

given :

initial velocity (u) = 0

final velocity (v) = 36 km/hr = 36*(5/18) = 10 m/s

time taken (t) = 10 minute = 10*60 = 600 seconds

to find : acceleration (a)

solution :

by using 1st equation of motion

v = u+at

10 = 0+a*600

a = 10/600

a = 0.0167 m/s²

so, the acceleration of train will be 0.0167 m/s²

Answer 2

ANSWER
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$initial \: velocity(u) \: = \: 0 \\ \\ final \: velocity(v) \: = \: 36 \: km \: {h}^{ - 1} \\ \\ since \: the \: final \: velocity \: is \: not \\ \: in \: its \: s.i. \: unit \: \\ we \: will \: change \: km \: {h}^{ - 1} \: into \: m \: {s}^{ - 1} \\ \\ 36 \: km {h}^{ - 1} \: = \: \frac{36 \times 1000}{60 \times 60} = \: 10 \: m \: {s}^{ - 1} \\ \\ \: time \: (t) \: = \: 10 \: mins \\ \\ since \: the \: time \: is \: not \: in \: s.i. \: unit \: \\ we \: will \: change \: it \: into \: seconds \\ \\ 10 \: mins = \: 10 \times 60 = \: 600 \: sec \\ \\ acceleration \: = \: \frac{rate \: f \: change \: of \: velocity}{time} \\ \\ a \: = \: \frac{ \: final \: velocity \: - \: initial \: velocity}{time} \\ \\ \\ \\ a \: = \: \frac{ \: 10 - 0}{600} \\ \\ a \: = \: \frac{10}{600} \\ \\ a \: = \: 0.017 \: m \: {s}^{ - 2} \: ans$


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