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A truck starts from rest and accelerates uniformly at 2.0 ms−2. At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What is the magnitude of acceleration (in ms−2 ) of the stone at t = 11s? (Neglect air resistance.)
  • a)
    8
  • b)
    14
  • c)
    10.0
  • d)
    12
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
A truck starts from rest and accelerates uniformly at 2.0 ms−2. ...
Initial velocity of the truck, u = 0
Acceleration, a = 2 m/s2
Time, t = 10 s
As per the first equation of motion, final velocity is given as:
v = u + at
⇒ 0 + 2 X 10 = 20 m/s
The final velocity of the truck and hence, of the stone is 20 m/s.
When the stone is dropped, its initial velocity is the same as the velocity of the truck at the time of release. The stone's motion can be broken into two components:
  1. Horizontal motion: The stone has the same horizontal velocity as the truck when dropped, which is 20 m/s. Since there is no force acting horizontally (neglecting air resistance), its horizontal velocity remains constant at 20m/s.
  2. Vertical motion: The stone experiences a downward acceleration due to gravity, which is g=9.8 m/s2.
Thus, the total acceleration of the stone at any point is the combination of:
  • The constant horizontal acceleration (which is 0 for the stone).
  • The vertical acceleration due to gravity, which is 9.8 m/s2 downward.
Therefore, at t=11 s, the total acceleration of the stone is the magnitude of the vector sum of these two accelerations:
However, since the truck is moving horizontally with an acceleration of 2.0 m/s2, at t=11s, the stone will still have the same horizontal acceleration as the truck. The total acceleration of the stone is the combination of:
  • The acceleration of the truck horizontally, 2.0 m/s2, and
  • The gravitational acceleration vertically, 9.8 m/s2.
Thus, the total magnitude of the acceleration of the stone at t=11s is:
 
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A truck starts from rest and accelerates uniformly at 2.0 ms−2. At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What is the magnitude of acceleration (in ms−2 ) of the stone at t = 11s? (Neglect air resistance.)a)8b)14c)10.0d)12Correct answer is option 'C'. Can you explain this answer? for NEET 2026 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A truck starts from rest and accelerates uniformly at 2.0 ms−2. At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What is the magnitude of acceleration (in ms−2 ) of the stone at t = 11s? (Neglect air resistance.)a)8b)14c)10.0d)12Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for NEET 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A truck starts from rest and accelerates uniformly at 2.0 ms−2. At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What is the magnitude of acceleration (in ms−2 ) of the stone at t = 11s? (Neglect air resistance.)a)8b)14c)10.0d)12Correct answer is option 'C'. Can you explain this answer?.
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