Question

A freight train initially moving at +10m/s decelerates at a rate of -0.1m/s^2. How far does it move in 30 seconds?

Answer 1

• Initial velocity=u=10m/s
• Acceleration=-0.1m/s^2=a
• Time=t=30s

• Distance=s

Apply second equation of kinematics

`\\ \tt\longmapsto s=ut+(1)/(2)at^2`

`\\ \tt\longmapsto s=10(30)+(1)/(2)(-0.1)(30)^2`

`\\ \tt\longmapsto s=300+(-0.1)(450)`

`\\ \tt\longmapsto s=300-45`

`\\ \tt\longmapsto s=255m`

Answer 2

255 [m].

Step-by-step explanation:

1) the formula is (d - requred distance, V₀ - initial velocity, t - elapsed time, a - deceleration):

`d=V_0t+(at^2)/(2);`

2) according to the formula above the required distance is:

d=10*30-0.5*0.1*900=300-45=255 [m].