Question

a train is travelling at a speed of 90 kilometre per hour brakes are applied so as to produce a uniform acceleration of -0.5 meter per second square.find how far the train will go before it is brought to rest.​

Answer 1

Answer:

Given:

Initial speed = 90 km/hr

Retardation = 0.5 m/s²

To find:

Distance travelled before the train stops.

Conversion:

Speed is converted from km/hr to m/s

90 km/hr = 90 × (5/18) = 25 m/s

Calculation:

∴ v² = u² + 2as

=> 0² = 25² + 2 × (-0.5) × s

=> s = 625 metres.

So final answer is :

$\boxed{\sf{ \red{distance \: = 625 \: metres}}}$

Answer 2

$\huge{\underline{\underline{\red{\mathfrak{AnSwEr :}}}}}$

$\small{\underline{\blue{\sf{Given :}}}}$

• Initial Velocity (u) = 90 km/h = 25m/s
• Final Velocity (v) = 0 m/s
• Retardation = -0.5 m/s2

$\rule{200}{1}$

$\small{\underline{\green{\sf{Solution :}}}}$

$\large \star{\boxed{\sf{v^2 \: - \: u^2 \: = \: 2as}}} \\ \\ {\sf{\pink{\: \: \: \: \: \: \: \: \: \: \: \: \: \: Put \: Values}}} \\ \\ \implies {\sf{( 0 )^2 \: - \: ( 25) ^2 \: = \: 2(-0.5)(s)}} \\ \\ \implies {\sf{0 \: - \: 625 \: = \: -1 \: \times \: s}} \\ \\ \implies {\sf{\cancel{-}625 \: = \: \cancel{-}s}} \\ \\ \implies {\sf{625 \: = \: s}}$

∴ Distance Covered (S) = 625 m