A train starts from rest and moves with a constant acceleration of $2.0\mathrm{m}/{\mathrm{s}}^2$ for half a minute. The brakes are then applied and the train comes to rest in one minute. Find (a) the total distance moved by train, (b) the maximum speed attained by the train, and (c) the position(s) of the train at half the maximum speed.

Step 1: Given data

Acceleration, $a=2.0\mathrm{m}/{\mathrm{s}}^2$

Initial velocity, $\mathrm{u}=0$

Step 2: Formula used

Use equations of motion to calculate the final velocity and the total distance traveled.

Final velocity, $\mathrm{v}=\mathrm{u}+\mathrm{at}$

Distance traveled, $\mathrm{x}=\mathrm{ut}+\frac{1}{2}{\mathrm{at}}^2$

Step 3: In case of accelerated motion

Given that, $\mathrm{a}=2.0\mathrm{m}/{\mathrm{s}}^2$

Let the final velocity before applying breaks be ${\mathrm{v}}_1$.

we know that,

Initial velocity, ${\mathrm{u}}_1=0$

Time taken, ${\mathrm{t}}_1=\frac{1}{2}\times 60=30\mathrm{seconds}$

Now final velocity is given by,

${\mathrm{v}}_1={\mathrm{u}}_1+{\mathrm{at}}_1$

${\mathrm{v}}_1=0+2\times 30$

${\mathrm{v}}_1=60\mathrm{m}/\mathrm{s}$

After brakes are applied, the maximum speed attained is $60\mathrm{m}/{\mathrm{s}}^2$.

Distance traveled

$\mathrm{x}=u_1{t}_1+\frac{1}{2}{{\mathrm{at}}_1}^2$

$\mathrm{x}=0+\frac{1}{2}\times 2\times {30}^2$

$\mathrm{x}=900\mathrm{m}$

$\mathrm{x}=0.9\mathrm{km}$

The maximum speed attained is $60\mathrm{m}/{\mathrm{s}}^2$, So ,let the distance traveled at half the maximum speed $\left(30\mathrm{m}/\mathrm{s}\right)$ be $y$,

Now from the equation

${v_1}^2={u_1}^2+2ax$

${30}^2=0^0+2\times 2\times \mathrm{y}$

$\Rightarrow \mathrm{y}=\frac{900}{4}=225\mathrm{m}$ …… (1)

Hence, the distance from the starting point is $225\mathrm{m}$.

Step 4: In case of decelerated motion

Initial velocity, ${\mathrm{u}}_2=60\mathrm{m}/\mathrm{s}$

Final velocity, ${\mathrm{v}}_2=0\mathrm{m}/\mathrm{s}$

Time taken, ${\mathrm{t}}_2=60\mathrm{s}$

From the equation

${\mathrm{v}}_2={\mathrm{u}}_2+a{\mathrm{t}}_2$

$$\begin{gathered} a=\frac{0-60}{60} \\ =-1\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Distance traveled,

$\mathrm{x}=u_2{t}_2+\frac{1}{2}a{{\mathrm{t}}_2}^2$

$\mathrm{x}=60\times 60+\frac{1}{2}\times (-1)\times {60}^2$

$\mathrm{x}=3600-1800$

$\mathrm{x}=1800\mathrm{m}$

$\mathrm{x}=1.8\mathrm{km}$

$\mathrm{The}\mathrm{total}\mathrm{distance}\mathrm{covered}=0.9+1.8=2.7\mathrm{km}$.

Let the distance moved at half the speed in this part be z

From

${{\mathrm{v}}_2}^2={{\mathrm{u}}_2}^2+2az$

${30}^2={60}^2+2\times (-1)\times \mathrm{z}$

$2\mathrm{z}={60}^2-{30}^2$

$2\mathrm{z}=2700$

$\mathrm{z}=1350\mathrm{m}$

$z=1.35\mathrm{km}$ …… (2)

Step 5: Conclusion

(a) Total distance traveled by train is given as,

$$\begin{gathered} \mathrm{s}=\mathrm{x}+\mathrm{z} \\ =900+1800 \\ =2700\mathrm{m} \\ =2.7\mathrm{km} \end{gathered}$$

(b) The maximum speed attained by the train is $60\mathrm{m}/\mathrm{s}$.

(c) The position of the train at half the maximum speed is given as,

There will be two positions at half the maximum speed (From the point of start),

From the starting point, the position is $225\mathrm{m}$.

The position after deceleration is$=1.35\mathrm{km}+0.9\mathrm{km}=2.25\mathrm{km}$.