Question

A train starting from arailway station and moving with uniform acceleration attains a speed of 54km/h-1 in 12min find its acceleration

Answer 1

Answer:

0.0208 m/s² is the Acceleration.

Explanation:

Here,

It says that the train has been uniformly accelerated which means the track was straight. So,

Speed = Velocity

Velocity = 54 km/hr

Velocity in m/s = (54*5/18) m/s

= 15 m/s

Time = 12 mins or, (12*60) seconds

= 720 seconds

Acceleration = Velocity/Time

= (15/720) m/s²

= 0.0208 m/s²

Hope it helps.

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Acceleration=0.02\:m/s^{2}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Final \: velocity(v) = 54 \: km/h \\ \\ \tt: \implies Time(t) = 12\:min \\ \\ \red{\underline \bold{To \: Find:}} \\ \tt: \implies Acceleration = ?$

• According to given question :

$\tt \circ \: Final \: velocity = 54 \times \frac{ 5}{18} = 15 \: m/s \\ \\ \tt \circ \: Time = 12 \times 60 = 720\: sec \\ \\ \tt \circ \: Initial \: velocity = 0 \: ms \\ \\ \bold{As \: we \: know \:that} \\ \tt: \implies v = u + at \\ \\ \tt: \implies 15 = 0 + a \times 720 \\ \\ \tt: \implies \frac{15}{720} = a \\ \\ \green{\tt: \implies a =0.02 \: {m/s}^{2} }$

$\bold{\blue{Some \: related \: formula }} \\ \orange{\tt \circ \: v^{2} = u^{2}+ 2as} \\ \\ \orange{\tt \circ \:s = ut + \frac{1}{2} at}$