Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# A Train Starts from Rest and Moves with a Constant Acceleration of 2.0 M/S2 for Half a Minute. Find the Position(S) of the Train at Half the Maximum Speed. - Physics

Short Note

A train starts from rest and moves with a constant acceleration of 2.0 m/s2 for half a minute. The brakes are then applied and the train comes to rest in one minute. Find the position(s) of the train at half the maximum speed.

#### Solution

Initial velocity, u = 0
Acceleration, a = 2 m/s2
Let the final velocity be v before the brakes are applied.
Now,
= 30 s
u + at
v = 0 + 2 × 30
⇒ v = 60 m/s

Half the maximum speed $= \frac{60}{2} = 30 \text{ m } /s$

When the train is accelerating with an acceleration of 2 m/s2:
Distance,

$s = \frac{v^2 - u^2}{2a'}$
$= \frac{{30}^2 - 0^2}{2 \times 2}$
⇒ s = 225 m

When the train is decelerating with an acceleration of - 1 m/s2:

Distance,

$s = \frac{v^2 - u^2}{2a'}$
$= \frac{{30}^2 - {60}^2}{2\left( - 1 \right)}$
⇒ s = 1350 m
Position from the starting point = 900 + 1350 = 2250
= 2.25 km

Is there an error in this question or solution?

#### APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 3 Rest and Motion: Kinematics
Q 15.3 | Page 52