Q80P - Fundamentals Of Physics

Question

A train started from rest and moved with constant acceleration. At one time, it was travelling30 m/s , and 160 m farther on it was travelling 50 m/s. Calculate (a) the acceleration (b) the time required to travel 160 m mentioned (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.

Step-by-Step Solution

Answer

1. The acceleration of the train is 5m/s²
2. The time required to travel the 160 m mentioned is 4 s
3. The time required to attain the speed of 30 m/s is 6 sec
4. The distance moved from rest to the time the train had a speed of 
5. The graph of x vs. t and v vs t

Step 1: Given information

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

Step 2: To understand concept of kinematics

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,

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

Step 3 a): Calculation for acceleration of train

According to the kinematic equations, 

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

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

Step 4 b): Calculation for time required to travel mentioned distance

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,

$t=4 s$

So, 4 sec are required to travel 160 m.

Step 5 c): Calculation for time to reach velocity of 30 m/s

According to the kinematic equations,

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

Step 6 d): Calculation for distance traveled to reach velocity of 30 m/s from rest

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 .

Step 7 e): Graphical representation of the given situation

                                                                       Graph of v vs t