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Motion in A Straight Line Test - 15

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Motion in A Straight Line Test - 15
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  • Question 1
    1 / -0
    When a ball is thrown vertically upwards, it reaches a maximum height of 5m. The initial velocity of the ball was ?
    Solution
    Answer is C.

    The equation is given as $${ v }^{ 2 }-{ u }^{ 2 }=2as$$.
    where,
    v = final velocity, 
    u = initial velocity, 
    a = acceleration due to gravity, $$10\quad m/{ s }^{ 2 }$$, 
    s = h = height traveled.
    Therefore, we can find u with above info. In this case, v = 0 at the top of the path.
    $$0-{ u }^{ 2 }=2\times 10\times 5\quad =\quad 10\quad m/s.$$
    Hence, the initial velocity of the ball was 10 m/s.
  • Question 2
    1 / -0
    A body will have uniform acceleration if its
    Solution
    If the velocity of an object changes at a uniform rate, then the acceleration that causes the change in velocity is called uniform acceleration or constant acceleration.
    For example, the force of gravity imparts an acceleration uniformly which is called acceleration due to gravity.
    Hence,option B is correct.
  • Question 3
    1 / -0
    The body will speed up if ___________ .
    Solution
    A body will speed up if both velocity and acceleration are in same direction. If they are in opposite directions it results in slowing down the motion. And if they are perpendicular then there will be no effect on magnitude of velocity.
  • Question 4
    1 / -0
    A stone falls from a balloon that is descending at a uniform rate of $$12ms^{-1}$$. The displacement of the stone from the point of release after 10sec is :-
    Solution
    $$S=12\times 10+\dfrac { 1 }{ 2 } \times 9.8\times { 10 }^{ 2 }\\ \ \ =120+490\\ \ \ =610\ m$$
  • Question 5
    1 / -0
    Which of the following is/are equations of uniformly accelerated motion relating the initial velocity (u), final velocity (v), time (t), acceleration (a) and displacement (S).
    Solution
    The equations of motion relate to the following five quantities:
    u - initial velocity
    v - final velocity
    a - acceleration
    t - time
    s - displacement
    Of the above u, v, a, and s are vector quantities. As such, remember to make vectors going in one direction positive and vectors in the opposite direction negative.
    Time (t) is a scalar quantity.
    If any three of the five quantities are known then the other two may be calculated using the following three equations:
    • $$v=u+at$$
    • $$s=ut+\frac { 1 }{ 2 } at^{ 2 }$$
    • $${ v }^{ 2 }{ =u }^{ 2 }+2as$$
    The equations of motion can only be used for an object travelling in a straight line with a constant acceleration.
  • Question 6
    1 / -0
    When an object is accelerated:
    Solution
    Acceleration is the time rate of change of velocity.
  • Question 7
    1 / -0

    Directions For Questions

    A car is moving on a straight road with constant acceleration. The initial velocity of the car is 5 m/s and after 10 seconds its velocity becomes 25 m/s. 

    ...view full instructions

    (a) Find the acceleration of the car
    Solution
    Given:
    Initial velocity $$u=5\ m/s$$
    Final velocity $$v=25\ m/s$$
    Time $$t=10\ s$$
    From the equation of motion $$v=u+at$$ we get,
    acceleration $$a=\dfrac{v-u}{t}=\dfrac{25-5}{10}=2\ m/s^2$$
  • Question 8
    1 / -0
    A particle experiences constant acceleration for $$20  s$$ after starting from rest. If it travels a distance $${X}_{1}$$, in the first $$10  s$$ and distance $${X}_{2}$$, in the remaining $$10  s$$, then which of the following is true ?
    Solution
     from eq $$s=ut+(1/2)a(t^2)$$

     initial velocity $$u=0 m/sec$$          (particle starts from rest)

    $$x_1=(1/2)a(t^2)$$

    $$x_1=(1/2)a(10^2)$$

    $$x_1=50a$$ ------------------(1)

    now initial velocity for $$x_2$$ is the final velocity of $$x_1$$

    let final velocity of $$x_1$$=initial velocity of $$x_2=v$$

    so, $$v=u +at$$

      $$v=at$$

     $$ v=10a$$  (since u=0 and t=10)

    now $$x_2=ut+(1/2)a(t^2)$$

    $$x_2=(10a)*10+(1/2)a(10^2)$$

    $$x_2=100a+50a$$

    $$x_2=150a$$ -------------------------(2)

    now

    $$\dfrac{(2)}{(1)}=\dfrac{x_1}{x_2}$$

    $$\dfrac{x_1}{x_2}=\dfrac{50a}{150a}$$

    $$3x_1=x_2$$
  • Question 9
    1 / -0
    Velocity-time graph of a body with uniform velocity is a straight line 
    Solution
    Velocity-time graph of an object moving with uniform velocity. The slope of a Velocity–time graph of an object moving in rectilinear motion with uniform velocity is straight line and parallel to x-axis when velocity is taken along y-axis and time is taken along x-axis.
  • Question 10
    1 / -0
    Area under a speed-time graph gives :
    Solution
    The area under the velocity-time graph gives the distance traveled by a moving object during that time interval.
    While the area under the velocity-time graph gives displacement of the particle.
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