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Motion Test - 34

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Motion Test - 34
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  • Question 1
    1 / -0
    A boy is whirling a stone tied with a string in a horizontal circular path. If the string breaks, the stone
    Solution
    When the boy is whirling a stone tied with a string in a  circular path, the centripetal force acts towards the center of the circular path. However, when the string breaks the centripetal force ceases and there is no force acting. Thus, the stone as per Newtons first law will continue in a straight line and fly off along the tangent to the circular path.
  • Question 2
    1 / -0
    An electron starting from rest has a velocity that increases linearly with time,i.e. v=kt where $$k=2\ m/s^2$$. The distance covered in the first three seconds will be
    Solution
    From the given information , v$$= kt $$
    Acceleration of the body 'a' $$= k = 2m/s^2 $$
    Using the equations of motion
    $$s = ut + \frac{1}{2}at^2 $$
    u = 0
    $$s = 2 \times (1/2) \times 3^2 $$
    $$s = 9m $$
  • Question 3
    1 / -0
    The velocity of a body at any instant is 10 m/s. After 5 sec, velocity of the particle is 20 m/s. The velocity at 3 seconds before is
    (assume uniform accelaration)
    Solution
    $$v_i = 10\ m/s $$ and $$v_f = 20\ m/s $$
    Acceleration of the body $$= \dfrac{20 - 10}{5} m/s^2 $$
    $$a = 2\ m/s^2 $$
    Velocity of the body at $$t= -3\ sec$$
    $$ v = u + at$$
    $$v = 10 - 2 \times 3 $$
    $$v = 4\ m/s $$
  • Question 4
    1 / -0

    In the velocity-time graph, AB shows that the body has

    Solution

    $$\textbf{Hint: }$$

    The slope of the curve in a velocity – time graph gives the acceleration.

    $$\textbf{Step1:Slope of velocity-time graph}$$

    Slope of velocity – time graph gives acceleration as shown below:

    $$slope=\dfrac{\Delta v}{\Delta t}=a$$

    $$\textbf{Step2:Conclusion from graph}$$

    According to the graph of the question, the initial velocity of the body at time $$t=0$$ is $$OA$$. The slope of the graph is constant but negative.

    Acceleration is a vector quantity having both magnitude and direction. Increase in velocity with respect to time is called acceleration whereas decrease in velocity with respect to time is called retardation. In this graph, the velocity is decreasing at a constant rate with respect to time. Hence, the body shows uniform retardation.

     

    The correct answer is option (D).

  • Question 5
    1 / -0
    Two objects of masses $$m_{1}$$ and $$m_{2}$$ having the same size are dropped simultaneously from heights $$h_{1}$$ and $$h_{2}$$ respectively. Find out the ratio of time they would take in reaching the ground.
    Solution
    Gravitational acceleration of the objects will not depend on their mass or size and hence it will remain same for all the objects. 
    $$h_1=u_1t_1+\dfrac{1}{2}at_1^2$$
    $$h_2=u_2t_2+\dfrac{1}{2}at_2^2$$
    $$u_1=u_2=0$$
    $$ \dfrac{t_1}{t_2} = (\dfrac {h_1}{h_2})^{\dfrac {1}{2}} $$
  • Question 6
    1 / -0
    A girl walks along a straight path to drop a letter in the letterbox and
    comes back to her initial position. Her displacement-time graph is shown in the figure. Find the average velocity.

    Solution
    $$\text{Average velocity} = \dfrac{\text{Total Displacement}}{\text{Total Time}}$$
    Since her displacement is $$zero$$
    Average velocity$$=0/100=0\ m/s$$
  • Question 7
    1 / -0
    Rain is falling vertically with a speed of 1.7 m/s. A girl is walking with speed of 1.0 m/s in the N - E (north-east) direction. To shield herself she holds her umbrella making an approximate angle $$\theta $$ with the vertical in a certain direction. Then


  • Question 8
    1 / -0
    A particle is moving in a circular path of radius r. Its displacement after moving through half the circle would be :
    Solution
    • Displacement is the minimum distance between the initial position and final position. 
    • After moving half circle (A to B) displacement will be AB 
    • AB is diameter i.e. $$2r$$
    • Option C is correct.

  • Question 9
    1 / -0
    A body is thrown vertically upwards and rises to a height of 10 m. The velocity with which the body was thrown upwards is($$g=9.8\ m/s^2$$)
    Solution
    Using the formula 
    $$v^2 = u^2 - 2gh $$
    At maximum height , v = 0
    $$u^2 = 2gH $$
    $$u^2 = 2 \times 9.8 \times 10 $$
    $$u^2 = 196 $$
    $$u = 14 m/s $$
  • Question 10
    1 / -0
    A body starting from rest in travelling with an acceleration of $$6m/s^2$$. Find the distance traveled by it in $$6^{th}$$ second.
    Solution
    Given,
    $$n=6$$
    $$u=0m/s$$
    $$a=6m/s^2$$
    The distance travelled by the body in $$n^{th}$$ second
    $$S_n=u+\dfrac{a}{2}(2n-1)$$
    $$s_6=0+\dfrac{6}{2}(2\times 6-1)=33mts$$
    The correct option is D.
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