91Ó°ÊÓ

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

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,

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

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