Question

A train runs at a speed of 120km/hr ina curved track of radius 900m the application of brakes suddenly causes the train to slow down at a constant rate after 6seconds the speed has been reduced to 92km/hr determine the acceleration immediately after the brake is applied​

Answer 1

Answer:

Explanation:

v = u +at;

92 = 120 + a( 6/3600)

so a = - 28*3600/6 =  -16800 km/h^2

negative sign shows retardation

changing this in m/s2

a = 16800 *1000/(3600*3600) = 1.296 m/s^2

Answer 2

Answer:

The acceleration after the brake is applied is 1.297 m/s2

Explanation:

Given Initial velocity (U) = 120km/hr

changing it into meter/sec = (120× 1000)/(60×60)

= 33.34m/sec.

Final velocity (V) = 92Km /hr

changing it into m/sec = (92×1000)/(60×60)

= 25.55 m/sec

Time taken = 6 seconds

$v - u = at$

= 25.55 - 33.34 = a × 6

= -7.78 = a × 6

= a = -7.78/6

= a = -1. 297 m

here the negative sign indicates deceleration.

So the acceleration after break is applied is 1.297m/s2.

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