Question

A train is travelling at a speed of 90km/hr. Brakes are applied so as to produce a uniform acceleration of - 0.5m second squared. Find how far the train will go before it is brought to rest.

Answer 1

$Hey \: friend \: here \: is \: your \: answer.$

$Speed \: of \: the\: train$ = 90km/hr
$Acceleration =$-0.5m/s²
$We \:know \:velocity \:is\: speed \:in\: a \:direction \:so \\INITIAL \:VELOCITY \:(u)$= 90km/hr
= 90 km / hr *1000 /3600
= 900m/36s
= 25 m / s
$FINAL\: VELOCITY\: (v)$= 0 m / s
$ACCELERATION\: (a)$ = -0.5 m / s²
$DISTANCE \:(s)$= ?
$Using \: 3rd \: Equation \: of \: motion$
2as = v² - u²
2 * ( -0.5 ) * s = ( 0 )² - ( 25 )²
-1s = 0 - 25
-s = -25
25 = s
$So \: Distance \: travelled \: is \: 25m.\\\\ Hope \: it \: helps.\\ Please \: mark \: my \: answer \: as \: the \\ BRILLIANIST \\ANSWER.$

Answer 2

Answer:625m

Explanation:

Initial velocity (u)= 90km/h

= 25m/s

Final velocity (v)= 0m/s

Acceleration (a)= -0.5m/s^2

Distance travelled (s)= ?

From 3rd equation of motion,

v^2 - u^2 = 2as

2as = v^2 - u^2

2× -0.5× s = 0^2 - 25^2

-1s = -25^2

1s = 25^2

s = 625m is the distance travelled