Source: High school physics (Chinese)
Problem
A train is traveling at a speed of $72$ km/h. As it approaches a station, it undergoes uniform deceleration and comes to a stop at the platform in $40$ s.
The distance from the platform where the train began its deceleration is given by the formula for displacement under constant acceleration:
$$\Delta x = \left(\frac{v_0 + v}{2}\right)t$$Substituting the values $v_0 = 20$ m/s, $v = 0$ m/s, and $t = 40$ s, the stopping distance is:
$$\Delta x = 400 \text{ m}$$The problem describes motion with uniform deceleration, which is a case of kinematics with constant acceleration.
First, convert the initial speed to SI units (m/s). Given the initial speed $v_0 = 72$ km/h.
$$v_0 = 72 \frac{\text{km}}{\text{h}} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}} = 20 \text{ m/s}$$The other given quantities are the final speed $v=0$ m/s (comes to a stop) and the time interval $t=40$ s.
[Q1] To find the distance from the platform where deceleration began, we need to calculate the displacement $\Delta x$ during this time. For an object undergoing constant acceleration, the displacement is the average velocity multiplied by the time interval.
The average velocity $\bar{v}$ is given by:
$$\bar{v} = \frac{v_0 + v}{2}$$The displacement $\Delta x$ is then:
$$\Delta x = \bar{v} t = \left(\frac{v_0 + v}{2}\right)t$$Substituting the known values into this expression:
$$\Delta x = \left(\frac{20 \text{ m/s} + 0 \text{ m/s}}{2}\right)(40 \text{ s})$$ $$\Delta x = (10 \text{ m/s})(40 \text{ s})$$ $$\Delta x = 400 \text{ m}$$The train began its deceleration 400 m from the platform.