A train started from rest and moved with constant acceleration. At one time, it was travelling30 m/s, and 160 mfarther on it was travelling 50 m/s. Calculate (a) the acceleration (b) the time required to travel 160 mmentioned (c) the time required to attain speed of 30 m/s(d) the distance moved from rest to the time the train had a speed of 30 m/s(e) Graph x vs t and v vs t for the train, from rest.

$$\begin{gathered} v_f=50\mathrm{m}/\mathrm{s} \\ {v}_0=30\mathrm{m}/\mathrm{s} \\ ∆x=160\mathrm{m} \end{gathered}$$

Kinematics is the studies of how a system of bodies moves without taking into account the forces or potential fields that influence the motion. The equations which are used in the study are known as kinematic equations of motion.

Formula:

The velocity is given by,

$$\begin{gathered} v_2^f=v_0^2+2a\left(∆x\right) \\ {v}_f=v_0+at \end{gathered}$$

The displacement is given by,

$∆x=\left(v_0\right)\left(t\right)+\frac{1}{2}a{t}^2$

According to the kinematic equations,

$$\begin{gathered} v_f^2=v_0^2+2a\left(∆x\right) \\ {50}^2={30}^2+\left(2\times a\times 160\right) \\ a=5m/s^2 \end{gathered}$$

So, the acceleration of the train would be$a=5m/s^2$

According to the newton’s second kinematic equation,

$$\begin{gathered} ∆x=\left(v_0\right)\left(t\right)+\frac{1}{2}a{t}^2 \\ 160=30t+\frac{1}{2}\left(5\right)t^2 \\ 160=30t+2.5t^2 \end{gathered}$$

Solving the quadratic equation we got,

$t=4s$

So, 4 sec are required to travel 160 m.

According to the kinematic equations,

$$\begin{gathered} v_f=v_0+a \\ 30=0+5 \\ t=6sec \end{gathered}$$

According to the newton’s second kinematic equation,

$$\begin{gathered} ∆x=\left(v_0\right)\left(t\right)+\frac{1}{2}a{t}^2 \\ x=\frac{1}{2}\left(5\right){\left(6\right)}^2 \\ x=60 \end{gathered}$$

So, to reach the velocity of 30 m/s from rest the train should travel a distance for 90m .

Graph of v vs t