A train starting from rest and Moving with a uniform acceleration attains a speed of 108 km per hour in 2.5 minutes. What will be the acceleration of the train and how much distance would it travel in this time ?

To solve this question, out of the equations of this type of motion, minimum how many of them are to be used ?

This question was previously asked in

Official Paper 5: OTET 2017 Paper 2 (Mathematics & Science)

Option 3 : Any two

English Pedagogy - 1

1342

15 Questions
15 Marks
15 Mins

**Concept:**

**Equations of Motion**

The equations of motion establish the relationship between acceleration, time, distance, initial speed, and final speed for a body moving in a straight line with uniform acceleration.

The equations are

v = u + at - (1)

\(s = ut + \frac{1}{2}at^2\) - (2)

v^{2} = u^{2} + 2as - (3)

v is final velocity, u is initial velocity, t is time, a is acceleration, s is the distance travelled.

**Calculation:**

Given, initial speed u = 0 (started from rest)

Final speed v = 108 km / hr = 108 × 5/18 m/s = 30 m/s

time t = 2.5 minutes = 2.5 × 60 sec = 150 second

acceleration a = ?

30 = 0 + a × 150

a = 30 /150 = 0.2 m/s^{2}

Now putting these in equation (2) we get

\(s = (0)t + \frac{1}{2}(0.2)(150)^2\)

s = 1125 ms

This can also be obtained using equation (3).

Number of Equations required

- In general, the number of equations formed is equal to the number of variables we need to find out.
- Here we have to find the acceleration of the train and distance.
- Rest Parameters like time, initial velocity, final velocity are given.
- So, two parameters are required and the rest parameters are given. So, we will be requiring two-equations.
- If we put, v, u, and t in the first equation, we can find a and if put the obtained value of a with other parameters in any of the other two equations (2 and 3) we can get s.
- So, we can solve the equation using any two equations.

__So, any two equations are required to solve the equation.__