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Q5:
A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 tonnes (Hint: 1 tonne = 1000 kg.)

Solution :

Step 1: Identifying Given Variables

Based on the kinematic parameters provided in the problem statement, we define the following variables:

  • Initial velocity ($u$): $0\text{ m/s}$ [Since the truck starts from rest]
  • Distance traveled ($s$): $400\text{ m}$
  • Time taken ($t$): $20\text{ s}$
  • Mass of the truck ($m$): $7\text{ tonnes} = 7 \times 1000\text{ kg} = 7000\text{ kg}$ [Conversion factor: $1\text{ tonne} = 1000\text{ kg}$]

Step 2: Calculating Acceleration ($a$)

To determine the constant acceleration, we utilize the second equation of motion [Per the Kinematic Equations of Uniformly Accelerated Motion]:

$s = ut + \frac{1}{2}at^2$

Substituting the known values into the equation:

$400 = (0 \times 20) + \frac{1}{2} \times a \times (20)^2$

$400 = 0 + \frac{1}{2} \times a \times 400$

$400 = 200a$

Solving for $a$:

$a = \frac{400}{200}$

$a = 2\text{ m/s}^2$

Step 3: Calculating Force ($F$)

To calculate the force, we apply Newton’s Second Law of Motion [Per Newton’s Second Law: $F = ma$]:

$F = m \times a$

$F = 7000\text{ kg} \times 2\text{ m/s}^2$

$F = 14000\text{ N}$

Visual Representation of Kinematic Motion

Truck (m=7000kg) Displacement (s = 400m) a = 2 m/s²

Summary Table of Calculations

Parameter Calculated Value
Acceleration ($a$) $2\text{ m/s}^2$
Force ($F$) $14000\text{ N}$

Final Solution: The acceleration of the truck is $2\text{ m/s}^2$ and the force acting on it is $14000\text{ N}$.


More Questions from Class 9 Science Force and Laws of Motion Exercises


CBSE Solutions for Class 9 Science Force and Laws of Motion


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