Self Studies
Self Studies
Self Studies
Self Studies
Selfstudy
Selfstudy

Motion Test - 53

Result Self Studies

Motion Test - 53
  • Score

    -

    out of -
  • Rank

    -

    out of -
TIME Taken - -
Self Studies

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Self Studies Self Studies
Weekly Quiz Competition
  • Question 1
    1 / -0
    With what speed should a bus travel so that it can cover a distance of 10 km in 5 min?
    Solution
    $$60$$ $$min$$ = $$1hour$$
    $$5$$ $$min$$ = $$\dfrac { 1 }{ 60 } \times 5=\dfrac { 1 }{ 12 } h$$

    $$Speed$$ = $$\dfrac { Distance }{ Time } $$
    $$Distance=10km$$
    $$time=\dfrac{1}{12}h$$

    $$speed$$= $$\dfrac { 10 }{ \dfrac { 1 }{ 12 }  } $$ = $$10\quad \times \quad 12$$
               = $$120$$ $$km/h$$
  • Question 2
    1 / -0
    A man runs $$20 \  km$$ in $$30 \ min.$$ What is his speed?
    Solution
    Given:
    Distance $$= 20 \ km$$
    Time $$=30 \ min =\dfrac{30}{60}hr= \dfrac { 1 }{ 2 } hr$$

    $$\therefore$$ Speed = $$\dfrac { Distance }{ time } $$ = $$\dfrac { 20 \ km}{ \left( \dfrac { 1 }{ 2 }hr \right) } $$
    $$\Rightarrow$$ Speed $$=40 \ km/hr$$

    Option (C) is correct.
  • Question 3
    1 / -0
    The displacement of a body starting from rest, and moving with an acceleration of $$1 \ { ms }^{ -2 }$$ at the end of 5 s is ____ m.
    Solution
    Given :   
    $$a = 1 \ m/s^2$$       $$t= 5 \ s$$
    Initial speed of the body  $$u = 0   \ m/s$$
    Displacement of the body can be calculated using 2nd equation of motion
    $$S= ut+\dfrac{1}{2}at^2$$
    $$\Rightarrow S = 0+\dfrac{1}{2}\times 1\times 5^2 = 12.5 \ m$$
  • Question 4
    1 / -0
    The ratio of the heights from which two bodies are dropped is 3 : 5 respectively. The ratio of their final velocities is
    Solution
    Initial velocity   $$u = 0$$
    Using   $$v^2-u^2 = 2gS$$
    Or   $$v^2 - 0 = 2gS$$
    $$\implies \ v = \sqrt{2gS}$$
    $$\implies \ v_1:v_2 = \sqrt{S_1}:\sqrt{S_2} = \sqrt{3}:\sqrt{5}$$
  • Question 5
    1 / -0
    If a ball thrown vertically up attains a maximum height of $$80\ m,$$ its initial speed is $$\displaystyle ( g = { 10\ ms }^{ -2 } )$$.
    Solution
    Maximum height attained 
    $$H = 80 \ m$$
    Initial speed  $$u = \sqrt{2gH} $$
    The final speed will be zero because at the highest point it changes its direction.
    $$\implies \ u  = \sqrt{2\times 10\times 80} = 40 \ m/s$$
  • Question 6
    1 / -0
    The velocity of a retarding body changes from $$\displaystyle { 90\ ms }^{ -1 }$$ to $$36\ ms^{ -1 }$$ in $$9$$ seconds. Find the retardation of a body?
    Solution
    Initial velocity  $$u = 90 \ ms^{-1}$$
    Final velocity  $$v = 36 \ ms^{-1}$$
    Time,  $$t = 9 \ s$$
    Using first equation of motion:
    $$v = u+at$$
    $$\Rightarrow 36 = 90+a\times 9$$
    $$\Rightarrow a = -6 \ m/s^2$$
    Thus retardation is  $$6 \ m/s^2$$
  • Question 7
    1 / -0
    The final velocity of a body starting from rest and moving with an acceleration $$\displaystyle 2  \ { ms }^{ -2 }$$ and covering a distance of 10 m is _____ $$\displaystyle { ms }^{ -1 }$$.
    Solution
    Initial velocity   $$u = 0 \ m/s$$
    Let final velocity be $$v$$.
    Acceleration  $$a = 2 \ m/s^2$$
    Distance covered  $$S = 10 \ m$$
    Using   $$v^2 - u^2 = 2aS$$
    $$\therefore$$   $$v^2 - 0 = 2\times 2\times 10$$
    $$\implies \ v = \sqrt{40} \ m/s$$
  • Question 8
    1 / -0
    Which of the following formulae is correct?
    Solution
    Speed of an object is defined as distance moved per unit time.It denotes how fast or how slow the object is moving.It is given by
    $$Speed=\dfrac{Distance}{time}$$
  • Question 9
    1 / -0
    Calculate the distance travelled by a man walking at a speed of 5 km/hr in 36 minutes.
    Solution
    Given s = 5km/hr = 5 $$\displaystyle \times \frac{5}{18}m/s$$
    t = 36 minutes = 36 $$\displaystyle \times $$ 60 sec
    $$\displaystyle \therefore $$ distance travelled = s $$\displaystyle \times $$ t
               = $$\displaystyle \frac{25}{18}\times 36\times 60 = 3000  m$$
              = $$\displaystyle \frac{3000}{1000}km = 3 km $$
  • Question 10
    1 / -0
    A car is travelling at a certain speed. Its final velocity reaches 40 m/s in 10 seconds accelerating at the rate of 2.5 m/s$$\displaystyle ^{2}$$. Find the initial velocity of the car.
    Solution
    Given,
    Time $$(t) = 10$$ seconds
    Accelerartion $$(a) = 2.5$$ m/s$$\displaystyle ^{2}$$
    We know $$v = u + at$$
    $$\displaystyle \Rightarrow 40 m/s = u + 2.5\times 10  \ {m}/{s^{2}}$$

    $$\displaystyle \Rightarrow 40 m/s = u + 25\ {m}/{s}$$

    $$\Rightarrow u = (40 - 25)\ \  m/s$$ 

          $$u = 15 m/s$$ 

    $$\displaystyle \therefore $$ initial velocity of the car $$= 15$$ m/s
Self Studies
User
Question Analysis

  • Correct -

  • Wrong -

  • Skipped -

My Perfomance
  • Score

    -

    out of -
  • Rank

    -

    out of -
Re-Attempt Weekly Quiz Competition
Self Studies Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Self Studies
Click on Allow to receive notifications
Allow Notification
Self Studies
Self Studies Self Studies
To enable notifications follow this 2 steps:
  • First Click on Secure Icon Self Studies
  • Second click on the toggle icon
Allow Notification
Get latest Exam Updates & FREE Study Material Alerts!
Self Studies ×
Open Now