Question

A train starts from rest and accelerates uniformly until it has traveled 5.6 km and acquired a forward velocity of The train then moves at a constant velocity of for 420 s. The train then slows down uniformly at until it is brought to a halt.The acceleration during the first5.6 km of travel is closest to:

a) 0.19 m/s2
b) 0.14 m/s2
c) 0.16 m/s2
d) 0.20 m/s2
e) 0.17 m/s2

Answer 1

c) 0.16 m/s2

Step-by-step explanation:

The computation of the acceleration during the first km of travel is shown below

Given that

Final velocity = v = 42 m/s

Initial velocity = u = 0 m/s

Distance = 5.6km

Based on the above information, we need to apply the following formula

As we know that

`v^2 - u^2 = 2as`

`a = (v^2 - u^2)/(2s)`

`= (42^2 - 0^2)/(2 * 5.6 * (1000\ m)/(1\ km) )`

= 0.1575 m/s ^2

hence, the correct option is c. 0.16 m.s^2