Back to Directory
NEET PHYSICSEasy

A body is moving with velocity 30 m/s30 \text{ m/s} towards east. After 10 s10 \text{ s} its velocity becomes 40 m/s40 \text{ m/s} towards north. The average acceleration of the body is:

A

7 m/s²

B

√7 m/s²

C

5 m/s²

D

1 m/s²

Step-by-Step Solution

  1. Vector Representation: Let East be along the positive x-axis (i^\hat{i}) and North be along the positive y-axis (j^\hat{j}).
  • Initial velocity: v⃗1=30i^ m/s\vec{v}_1 = 30\hat{i} \text{ m/s}
  • Final velocity: v⃗2=40j^ m/s\vec{v}_2 = 40\hat{j} \text{ m/s}
  1. Change in Velocity: Δv⃗=v⃗2−v⃗1=40j^−30i^=−30i^+40j^ m/s\Delta \vec{v} = \vec{v}_2 - \vec{v}_1 = 40\hat{j} - 30\hat{i} = -30\hat{i} + 40\hat{j} \text{ m/s}.
  2. Average Acceleration Formula: Average acceleration is defined as the change in velocity divided by the time interval . a⃗av=Δv⃗Δt=−30i^+40j^10=−3i^+4j^ m/s2\vec{a}_{av} = \frac{\Delta \vec{v}}{\Delta t} = \frac{-30\hat{i} + 40\hat{j}}{10} = -3\hat{i} + 4\hat{j} \text{ m/s}^2
  3. Magnitude: The magnitude of the average acceleration is: ∣a⃗av∣=(−3)2+(4)2=9+16=25=5 m/s2|\vec{a}_{av}| = \sqrt{(-3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \text{ m/s}^2
Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started
Solved: PHYSICS Question for NEET | Sushrut