Question

A body is moving with velocity $30\, ms$ towards east After 10 s its velocity becomes $40 \,ms$ towards north. The average acceleration of the body is

• A. $7\, ms^{-1}$
• B. $\sqrt{7}\,ms^{-1}$
• C. $5\,ms^{-1}$
• D. $1 \,ms^{-2}$

Answer 1

The Correct Option is C

To calculate the average acceleration of the body, we need to assess the change in velocity and the time over which this change occurs.

The initial velocity of the body is 30 \, \text{m/s} towards east, which we can represent as a vector \mathbf{v_i} = 30 \, \hat{i}.

The final velocity after 10 seconds is 40 \, \text{m/s} towards north, represented as \mathbf{v_f} = 40 \, \hat{j}.

First, calculate the change in velocity:

• Change in velocity, \Delta \mathbf{v} = \mathbf{v_f} - \mathbf{v_i}
• \Delta \mathbf{v} = 40 \, \hat{j} - 30 \, \hat{i} = -30 \, \hat{i} + 40 \, \hat{j}

Now calculate the magnitude of the change in velocity using the Pythagorean theorem:

• |\Delta \mathbf{v}| = \sqrt{(-30)^2 + (40)^2}
• |\Delta \mathbf{v}| = \sqrt{900 + 1600}
• |\Delta \mathbf{v}| = \sqrt{2500}
• |\Delta \mathbf{v}| = 50 \, \text{m/s}

The time interval \Delta t over which this change occurs is 10 s.

Average acceleration \mathbf{a_{avg}} is calculated as:

• \mathbf{a_{avg}} = \frac{|\Delta \mathbf{v}|}{\Delta t}
• \mathbf{a_{avg}} = \frac{50 \, \text{m/s}}{10 \, \text{s}}
• \mathbf{a_{avg}} = 5 \, \text{m/s}^2

Therefore, the average acceleration of the body is 5 \, \text{m/s}^2.