Question

The time taken by a train to attain a velocity from 20 kmh-1 to 92 kmh-1 with a uniform acceleration of 2 ms-2 is

Answer 1

Answer:

Convert 20 km/h and 92 km/h in m/s

20*5/18 = 5.5m/s

92*5/18 = 25.5m/s

Using the formula for acceleration

2m/s^2 = 25.5-5.5/t

t = 20/2

t = 10s

Answer 2

Answer:

180s

Explanation:

Given

initial velocity =20km/h=20000/3600=100/18 m/s

final velocity = 92km/h=92000/3600=460/18 m/s

a=2m/s²

TO find=time

solution=

ACCELERATION=f(inal velocity -Initial velocity )/time

$2 = \frac{ \frac{460}{18} - \frac{100}{18} }{time} \\ 2 = \frac{360}{t } \\ \\ t = 360 \div 2 \\ t = 180$