A train is moving with a velocity of 400 m/s. With the application of brakes a retardation of $10 m / s^{2}$ is produced. Calculate the following :
(i) Time taken to come to rest
(ii) Braking distance [5 MARKS]

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Solution

Finding time : 2.5 Marks
Findinig distance : 2.5 Marks

(i) Given
$u = 400 m / s, a = - 10 m / s^{2}, v = 0$
$U s i n g e q u a t i o n, v = u + a t, w e g e t$
$0 = 400 + (- 10 \times t)$
$o r t = 40 s$

(ii) For calculating the distance travelled, we use equation,
$v^{2}$ = $u^{2} + 2 a S, w e g e t$
$0 = 400^{2} + (2 \times (- 10) \times S)$
$o r 20 S = 400 \times 400$
$o r S = 8000 m = 8 k m$

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A train is moving with a velocity of 400 m/s. With the application of brakes a retardation of 10m/s2 is produced. Calculate the following : (i) Time taken to come to rest (ii) Braking distance [5 MARKS]

Finding time : 2.5 Marks Findinig distance : 2.5 Marks (i) Given u=400 m/s,a=–10 m/s2, v=0 Using equation, v=u+at, we get 0=400+(–10×t) ort=40 s (ii) For calculating the distance travelled, we use equation, v2 = u2+2aS, we get 0=4002+(2×(–10)×S) or 20S=400×400 or S=8000 m=8 km

A train is moving with a velocity of 400 m/s. With the application of brakes a retardation of $10 m / s^{2}$ is produced. Calculate the following :
(i) Time taken to come to rest
(ii) Braking distance [5 MARKS]