Question

a train starting from a railway station and moving with uniform acceleration attains a speed of 40 km per hour in 10 minutes. find its acceleration

Answer 1

$\huge{\mathcal{Hi\: there!}}$

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Given :

Initial velocity ( u ) = 0 m / s.

Final velocity ( v ) = 40 km / hr

⇒ 40 × 5 / 18

⇒200 / 18

⇒100 / 9 m/s.

Time taken ( t ) = 10 mins = 600 secs.

Now,

We will use first equation of motion here ;

v = u + at

⇒ a = $\mathsf{ \frac{v - u}{t} }$
$\mathsf{⇒ a = \frac{v - u}{t} } \\ \\ \mathsf{⇒ a = \frac{1 \cancel{00}}{9} - 0 \times \frac{1}{6 \cancel{00}} } \\ \\ \mathsf{⇒ a = \frac{1}{9} \times \frac{1}{6} } \\ \\ \mathsf{⇒ a = \frac{1}{54} } \\ \\ \mathsf{ \boxed{ \bold{a = 0.018 \: \: per \: sq \: m}}}$

Hence, the acceleration is 0. 018 m / s².

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Thanks for the question !

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Answer 2

Answer:

→ Acceleration : 0.018 ms^-2

Step-by-step explanation:

Given,

Initial velocity (u) = 0

Final velocity (v) = 40 km/h = 11.11 m/s

Time (t) = 10 m = 600 s

[How we got final velocity → v = 40000/3600 = 11.11 m/s]

★ Acceleration : v - u/t

= (11.11 - 0) m/s / 600 s

= 0.018 m/s² or 0.018 ms^-2

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