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NEET PHYSICSEasy

An object with a mass 10 kg moves at a constant velocity of 10 m/s10 \text{ m/s}. A constant force then acts for 4 seconds on the object and gives it a speed of 2 m/s2 \text{ m/s} in the opposite direction. The acceleration produced in it is:

A

3 m/s23 \text{ m/s}^2

B

−3 m/s2-3 \text{ m/s}^2

C

0.3 m/s20.3 \text{ m/s}^2

D

−0.3 m/s2-0.3 \text{ m/s}^2

Step-by-Step Solution

  1. Sign Convention: Let the initial direction of motion be positive.
  • Initial velocity, u=+10 m/su = +10 \text{ m/s}.
  • Final velocity, v=−2 m/sv = -2 \text{ m/s} (since it is in the opposite direction).
  1. Time: The time interval, t=4 st = 4 \text{ s}.
  2. Formula: Acceleration (aa) is defined as the rate of change of velocity. a=v−uta = \frac{v - u}{t} [Source 34]
  3. Calculation: a=−2−104a = \frac{-2 - 10}{4} a=−124a = \frac{-12}{4} a=−3 m/s2a = -3 \text{ m/s}^2 The negative sign indicates that the acceleration is acting in the direction opposite to the initial motion (retardation followed by acceleration in the reverse direction).
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