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Motion Test - 49

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Motion Test - 49
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  • Question 1
    1 / -0
    A truck increases its speed from 10km/h10km/h to 50km/h50 km/h in 2020 seconds. Its acceleration is:
    Solution
    Given: Initial speed, u=10 km/h=10×518m/su=10\ km/h=10\times \dfrac{5}{18}m/s

                Final speed, v=50 km/h=50×518m/sv=50\ km/h=50\times \dfrac{5}{18}m/s

    Using equation of motion,

                Acceleration, a=vuta=\dfrac{v-u}{t}

                                          =518×(5010)20=\dfrac{\dfrac{5}{18}\times(50-10)}{20}

                                          =518×2=\dfrac{5}{18}\times 2

                                          =0.55 m/s2=0.55\ m/s^2
  • Question 2
    1 / -0
    A stone is thrown vertically upward with an initial velocity of 40ms140\, ms^{- 1}. Taking g=10ms2g\, =\, 10\, ms^{- 2}, the maximum displacement and distance covered by the stone on reaching the ground is 
    Solution
    Answer is B.

    Given that Initial velocity, u = 40 m/s, Final velocity, v = 0 m/s.
    Acceleration due to gravity, g = 10m/s210m/{ s }^{ 2 } (downward motion).
    Maximum height, s = H.
    As the body is thrown upward a = -g the relation v2=u22as{ v }^{ 2 }={ u }^{ 2 }-2as gives v2=u22aH{ v }^{ 2 }={ u }^{ 2 }-2aH, we have,
    H = u2v22g=40m/s2022(10m/s2)=160020=80m\dfrac { { u }^{ 2 }-{ v }^{ 2 } }{ 2g } =\dfrac { { 40m/s }^{ 2 }-{ 0 }^{ 2 } }{ 2(10m/{ s }^{ 2 }) } =\dfrac { 1600 }{ 20 } =80\quad m.
    If a stone is thrown vertically upward, it returns to its initial position after achieving maximum height.
    so, the net displacement = Difference of positions between initial and final positions = 0.
    Total distance covered = 80 m + 80 m = 160 m.
    Hence, the displacement is 0 and the total distance covered is 160 m.
  • Question 3
    1 / -0
    A moving bus is brought to rest within 30 seconds by applying brakes. What is its initial velocity, if the retardation due to brakes is 2 ms2ms^{-2}?
    Solution
    Given: 
    Time, t=30 sect=30\ sec
    Retardation, a=2 m/s2a=-2\ m/s^2
    Final velovity, v=0v=0
    From equation of motion
    v=u+at \Rightarrow v=u+at
    0=u2×30 \Rightarrow 0=u-2\times 30
    u=60 m/s\Rightarrow u=60 \ m/s
  • Question 4
    1 / -0
    A car is moving with a uniform velocity of 40 km h1^{-1} in straight line. Its acceleration after 1 hour is:
    Solution
    Answer is D.

    When the car is moving with uniform velocity, then there is nil acceleration in the motion of the car.
    Therefore, the total acceleration of the car is 0.
  • Question 5
    1 / -0
    A ball starts rolling on a horizontal surface with an initial velocity of 1 m/s.1 \ m/s. Due to friction, its velocity decreases at the rate of 0.1m/s2{0.1 m/s^2}. How much time will it take for the ball to stop?
    Solution
    In the given question, we have 
    u= 1 m/su= \ 1 \ m/s
    a=0.1 m/s2a = - 0.1 \ m/s^2 (negative sign indicates that the direction of acceleration is opposite to the initial velocity)
    As the ball finally stops,
    v=0v=0
    t=?t=?
    According to first equation of motion,
    v=u+atv = u + at
    0=1+(0.1)t\Rightarrow 0 = 1 + (-0.1)t
    1=0.1t\Rightarrow 1 = 0.1 t
    So t=10 sect=10 \ sec 
    Option C is correct.
  • Question 6
    1 / -0
    A vehicle moving at a speed of 15 m/s is stopped by applying brakes which produce a uniform acceleration of -0.5 m/s2m/s^2. The distance covered by the vehicle before it stops is:
    Solution
    Initial velocity or speed u=15m/su= 15m/s
    The acceleration a=0.5m/s2 a= -0.5m/s^2
    When vehicle stops its final velocity becomes zero v=0v=0
    By applying equation of motion 
    v2=u2+2aS{ v }^{ 2 }={ u }^{ 2 }+2aS
    0=2252×0.5SS=2252×0.5=225m0=225-2\times 0.5 S\\ S=\dfrac { 225 }{ 2\times 0.5 } =225 m

    Option D is correct.

  • Question 7
    1 / -0
    A boy running at an average speed of 4km/hr4 km/ hr reaches school from his home in 12 hr\dfrac{1}{2}\ hr. The distance of the school from his home is
    Solution
    Given:
    Average speed of the boy v=4 km/hv =4\ km/h
     Time taken t=12t =\dfrac{1}{2}hour
    Distance D=?D=?
      
     We know that  Speed(v)=Distance(D)Time(t)Speed(v)=\dfrac{Distance(D)}{Time(t)}
            Distance(D)=4×12\Rightarrow Distance(D)=4\times\dfrac{1}{2}
            Distance(D)=2 km\Rightarrow Distance(D)=2\ km
  • Question 8
    1 / -0
    When a bus driver travelling at a speed of 20m/s applied brakes and brings the bus to rest in 10 seconds, then retardation will be
    Solution
    In the given question 
    u=20m/s
    v=0m/s
    t=10s
    and a=?
    we know the formula v =u + at
     so putting the values we get 
    0 = 20 + a X 10
    - 20 = 10 a
    -2 = a 
    we get acceleration is -2 so retardation will be +2 m/ sec sq.
  • Question 9
    1 / -0
    AA and BB are arguing about uniform acceleration. AA states that acceleration means "the longer you go." BB states that acceleration means "the further you go." Who is right?
    Solution
    Acceleration of the object means how fast the object is changing its velocity or at what rate its velocity is changing. Acceleration does not mean distance covered by the object or time taken to cover that distance. So, none of them is right.
  • Question 10
    1 / -0
    A body has an acceleration of 4 ms2\displaystyle -4\ { ms }^{ -2 }. Then its retardation is equal to:
    Solution
    The change in the velocity of an object per unit time is known acceleration.
    Retardation is negative acceleration.
    Given,
    Acceleration, a=4 m/s2a=-4\ m/s^2
    Retardation =a=(4)=4 m/s2=-a=-(-4) =4\ m/s^2
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