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

Answer:

255 [m].

Explanation:

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

$d=V_0t+\frac{at^2}{2};$

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

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

Answer 2

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

• Distance=s

Apply second equation of kinematics

$\\ \tt\longmapsto s=ut+\dfrac{1}{2}at^2$

$\\ \tt\longmapsto s=10(30)+\dfrac{1}{2}(-0.1)(30)^2$

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

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

$\\ \tt\longmapsto s=255m$