Solved Examples on Rectilinear Motion

Example 1: A car accelerates uniformly from rest to a speed of 20 meters per second in 10 seconds. What is the acceleration of the car?

Solution:

Given, initial velocity (u) = 0 meters per second, final velocity (v) = 20 meters per second, time (t) = 10 seconds.

Using the formula for acceleration: Acceleration (a) = (final velocity – initial velocity) / time Acceleration

(a) = (20 m/s – 0 m/s) / 10 s

Acceleration (a) = 20 m/s / 10 s

Acceleration (a) = 2 meters per second squared

So, the acceleration of the car is 2 meters per second squared.

Example 2: A stone is thrown vertically upward with a velocity of 30 meters per second. How high does it go before it starts to fall back down? (Take acceleration due to gravity as 10 meters per second squared).

Solution:

Given, initial velocity (u) = 30 meters per second, acceleration due to gravity (g) = -10 meters per second squared (negative because it acts in the opposite direction to the motion).

Using the equation of motion: Final velocity (v) = 0 meters per second (at the highest point) v² = u² + 2as

So, 0 = (30 m/s)² + 2(-10 m/s²)s

0 = 900 -20s

s = 900 / 20

s = 45 meters

So, the stone reaches a height of 45 meters before it starts to fall back down.

Example 3: A train accelerates uniformly from 10 meters per second to 30 meters per second in 5 seconds. What is its acceleration?

Solution:

Given, initial velocity (u) = 10 meters per second, final velocity (v) = 30 meters per second, time (t) = 5 seconds.

Using the formula for acceleration:

Acceleration (a) = (final velocity – initial velocity) / time

Acceleration (a) = (30 m/s – 10 m/s) / 5 s

Acceleration (a) = 20 m/s / 5 s

Acceleration (a) = 4 meters per second squared

So, the acceleration of the train is 4 meters per second squared.

Example 4: A ball is dropped from a height of 20 meters. What is its velocity just before hitting the ground? (Take acceleration due to gravity as 10 meters per second squared)

Solution:

Given, initial velocity (u) = 0 meters per second, acceleration due to gravity (g) = 10 meters per second squared, displacement (s) = -20 meters (negative because it is downward).

Using the equation of motion: v² = u² + 2as

So, v² = 0² + 2(10 m/s²)(-20 m)

v² = 0 + (-400 m²/s²)

v² = -400 m²/s²

Since velocity cannot be negative, we take the magnitude: v = √(400 m²/s²) v = 20 meters per second

So, the velocity of the ball just before hitting the ground is 20 meters per second.

Rectilinear motion refers to the motion of an object along a straight line. In rectilinear motion, the object’s position changes with respect to time, but its path remains linear, following a single dimension.

This article discusses rectilinear motion, its characteristics, equations, types, laws, differences from linear motion, and frequently asked questions.