Kinematics Part-1 (one dimension) questions with solutions

Showing 1–10 of 487 questions

• MCQmediumKinematics Part-1 (one dimension)Instantaneous speed and instantaneous velocity

  For x = t⁴ - 8t² during 0 ≤ t ≤ 4 s, the ratio of average speed to magnitude of average velocity is:

  Options

  • Option A

    4/5

  • Option B

    5/4

  • Option C

    3/2

  • Option D

    2

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• Multi-correctmediumKinematics Part-1 (one dimension)Motion under constant acceleration

  For one-dimensional motion with constant acceleration, which statements are correct?

  Select all that apply

  • Option A

    Velocity is a linear function of time.

  • Option B

    Position is a quadratic function of time in general.

  • Option C

    v² is a linear function of position as long as acceleration remains constant.

  • Option D

    Signed displacements in successive equal time intervals form an arithmetic progression.

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• MatchdifficultKinematics Part-1 (one dimension)Ghraphical problem of kinematics

  Match Column-I with Column-II and choose the correct option.

  List 1

  1. A. Acceleration decreasing with time

  2. B. Velocity increasing with time

  3. C. Magnitude of acceleration increasing with time

  4. D. Body going farther away from the starting point with time

  List 2

  1. p.

     

  2. q.

     

  3. r.

     

  4. s.

     

  Options

  • Option A

    A-(r); B-(p,q); C-(s); D-(p,q,r,s)

  • Option B

    A-(r,s); B-(r,q); C-(p,q,r,s); D-(s)

  • Option C

    A-(r,q); B-(q); C-(s,r); D-(p,q,r,s)

  • Option D

    A-(r,s); B-(p,r); C-(p,q,r,s); D-(p,q)

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• MCQmediumKinematics Part-1 (one dimension)Ghraphical problem of kinematics

  The velocity time graph of a body moving in a straight line is shown in figure. The ratio of displacement to distance travelled by the body in time 0 to 10s is

  

  Options

  • Option A

    1:1

  • Option B

    1:4

  • Option C

    1:2

  • Option D

    1:3

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• MCQmediumKinematics Part-1 (one dimension)Motion under gravity

  From the top of a tower, a stone thrown upward reaches the ground in time t₁. A second stone thrown downward with the same speed reaches the ground in time t₂. A third stone released from rest reaches the ground in time t₃. Which relation is correct?

  Options

  • Option A

    t₃=(t₁+t₂)/2

  • Option B

    t₃=√(t₁t₂)

  • Option C

    1/t₃=1/t₁-1/t₂

  • Option D

    t₃²=t₁²-t₂²

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• MCQmediumKinematics Part-1 (one dimension)Ghraphical problem of kinematics

  Which option is correct for the object having straight-line motion represented by the graph?

  

  Options

  • Option A

    The object moves with constantly increasing velocity from O to A and then with constant velocity.

  • Option B

    Velocity of the object increases uniformly.

  • Option C

    Average velocity is zero.

  • Option D

    The graph shown is impossible.

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• MCQmediumKinematics Part-1 (one dimension)Acceleration and Avg acceleration

  A particle moves along the x-axis according to x = t⁴ - 8t³ + 18t². The velocity at the first positive instant when its acceleration becomes zero is:

  Options

  • Option A

    0 m s⁻¹

  • Option B

    8 m s⁻¹

  • Option C

    16 m s⁻¹

  • Option D

    24 m s⁻¹

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• MCQmediumKinematics Part-1 (one dimension)Motion under constant acceleration

  A particle moves with speed v=b√x along the positive x-axis. Assuming it is at origin at t=0, calculate its speed at time t=τ.

  Options

  • Option A

    b²τ/4

  • Option B

    b²τ/2

  • Option C

    b²τ

  • Option D

    b²τ/√2

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• MCQmediumKinematics Part-1 (one dimension)Distance,Displacement ,Avg speed and Avg velocity

  A particle moves continuously along a line. In every subinterval its average speed equals the magnitude of average velocity. Which statement is possible?

  Options

  • Option A

    It repeatedly reverses direction.

  • Option B

    Its speed varies but direction remains unchanged.

  • Option C

    Its velocity must be constant.

  • Option D

    Its acceleration must always be zero.

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• MCQmediumKinematics Part-1 (one dimension)Acceleration and Avg acceleration

  The motion of a body is given by dv/dt = 6.0-3v(t), where v is in m s⁻¹ and t in s. If the body is at rest at t=0, find the wrong option.

  Options

  • Option A

    The terminal speed is 2.0 m s⁻¹.

  • Option B

    The speed varies as v(t)=2(1-e⁻³ᵗ) m s⁻¹.

  • Option C

    The speed is 0.1 m s⁻¹ when the acceleration is half the initial value.

  • Option D

    The magnitude of initial acceleration is 6.0 m s⁻².

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