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Motion in A Straight Line Test - 27

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Motion in A Straight Line Test - 27
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  • Question 1
    1 / -0
    Engine of a vehicle can give it an acceleration of $$1\;m{ s }^{ -2 }$$ and its brakes can retard it at $$3\;m{ s }^{ -1 }$$. The minimum time in which the vehicle can make a journey between stations $$A$$ and $$B$$ having a distance of $$1200\;m$$ is
    Solution
    Let $$x$$ be the distance traveled until maximum velocity (V) and $$y$$ be the remaining distance
    we can write
    $$ { v }^{ 2 }-0=2.1.x\quad \Rightarrow \quad x=\dfrac { { v }^{ 2 } }{ 2 }$$
    Similarly
    $$0-{ v }^{ 2 }=-6y\quad \Rightarrow \quad y=\dfrac { { v }^{ 2 } }{ 6 } \\ x+y=1200=\dfrac { { 2v }^{ 2 } }{ 3 } \Rightarrow { v }=30\sqrt { 2 }$$
    Now, using equation $$ v=u+at$$
    we get
    $${ t }_{ 1 }=30\sqrt { 2 } \quad \& \quad { t }_{ 2 }=10\sqrt { 2 } \\ { t }_{ 1 }+{ t }_{ 2 }=40\sqrt { 2 } =56.5$$ s
  • Question 2
    1 / -0
    A soldier jumps out from an aeroplane with a parachute. After dropping through a distance of $$19.6\ m$$, he opens the parachute and decelerates at the rate of $$1\ m{ s }^{ -2 }$$. If he reaches the ground with a speed of $$4.6\ m{ s }^{ -1 }$$, how long was he in air?
    Solution
    Let $$ { t }_{ 1 }$$ be the time before opening of parachute.
    Using $$ h=ut+\dfrac { 1 }{ 2 } g{ t }^{ 2 }$$, we get, $$ 19.6=0+\dfrac { 1 }{ 2 } 9.8{ t }_{ 1 }^{ 2 }\quad \Rightarrow \quad t_1=2s$$
    Taking $$v_1$$ as velocity attained after falling through $$19.6\ m$$
    Using $$ { v }^{ 2 }={ u }^{ 2 }+2gh$$, we get $$ { v }_{ 1 }^{ 2 }=0+2\left( 9.8 \right) \left( 19.6 \right) \quad \Rightarrow { v }_{ 1 }=19.6\ m/s$$
    Again taking $$ { t }_{ 2 }$$ as time taken after opening of parachute 
    Using $$v=u+at, $$ we get  $$ 4.6=19.6-\left( 1 \right) { t }_{ 2 }\quad \Rightarrow { t }_{ 2 }=15\ s$$
    Total time $$ ={ t }_{ 1 }+{ t }_{ 2 }=2+15=17\ s$$ 
  • Question 3
    1 / -0
    An object may have
    (I) varying speed without having varying velocity.
    (II) varying velocity without having varying speed.
    (III) non-zero acceleration without having varying velocity.
    (IV) non-zero acceleration without having varying speed.
    Solution
    Speed is a scalar quantity while velocity is a vector quantity. 
    An object can have constant speed but velocity will change since velocity is a vector quantity. 
    Without having a varying speed non-zero acceleration.  Uniform circular motion is an example of options B and D
    Thus, B and D options are correct
  • Question 4
    1 / -0
    A body travels $$200\ cm$$ in the first two seconds and $$220\ cm$$ in the next four seconds. What will be the velocity at the end of $$7^{th}$$ second from the start?
    Solution
    $$ s= ut + \dfrac { 1 }{ 2 } a t^2$$
    For $$case\   1.$$
    $$ 200 = 2u + 2a$$
    $$ u + a = 100   ....(1)$$
    For $$case\   2.$$ .i.e. next 4 seconds, we have distance $$= 420\  cm$$ and time $$= 6\  s$$
    $$ 420 =6 u+18a$$
    $$ u + 3a= 70  ....(2) $$
    Solving equation $$1$$ and $$2,$$ we get, 
    $$u = 115\  cm/s$$ and $$a = -15\ cm/s^2$$
    Now by 
    $$ v=u+ at$$
    $$ v = 115 -15\times 7$$
    $$ v =10\  cm/s $$
    Hence answer is A.
  • Question 5
    1 / -0
    A stone is allowed to fall from the top of a tower and covers half the height of the tower in the last second of its journey. The time taken by the stone to reach the foot of the tower is
    Solution
    Let total distance traveled be $$h$$.
    Now, Velocity at midpoint, $$v=\sqrt { 2gs } =\sqrt { gh } $$

    Velocity at lowest point, $$V=v=\sqrt { 2gs } =\sqrt { 2gh } $$

    Now, $$V=v+gt$$    $$(t=1s)$$

    Now, $$\sqrt { 2gh } =\sqrt { gh } +g$$

    $$ \Rightarrow \sqrt { h } =\dfrac { \sqrt { g }  }{ \sqrt { 2 } -1 } $$

    Also, $$h=\dfrac { 1 }{ 2 } g{ t }^{ 2 }$$

    $$\Rightarrow { t }^{ 2 }-\dfrac { 2h }{ g } =0$$

    $$ t=\pm \sqrt { \dfrac { 2h }{ g }  } =\pm \dfrac { \sqrt { 2 }  }{ \sqrt { 2 } -1 } \\ t=2\pm \sqrt { 2 } $$
  • Question 6
    1 / -0
    An aeroplane drops a parachutist. After covering a distance of $$40\ m$$, he opens the parachute and retards at $$2\ m{ s }^{ -2}$$. If he reaches the ground with a speed of $$2\ m{ s }^{ -1 }$$, he remains in the air for about
    Solution
    Let $$ { t }_{ 1 }$$ be the time before opening of parachute.
    Using $$ h=ut+\dfrac { 1 }{ 2 } g{ t }^{ 2 }$$, we get, $$ 40=0+\dfrac { 1 }{ 2 } 9.8{ t }_{ 1 }^{ 2 }\quad \Rightarrow \quad { t }_{ 1 }=3s$$
    Taking $$ { v }_{ 1 }$$ as velocity attained after falling through $$40m$$
    Using $$ { v }^{ 2 }={ u }^{ 2 }+2gh$$, we get $$ { v }_{ 1 }^{ 2 }=0+2\left( 9.8 \right) \left( 40 \right) \quad \Rightarrow { v }_{ 1 }=28\ m/s$$
    Again taking $$ { t }_{ 2 }$$ as time taken after opening of parachute 
    Using $$v=u+at, $$ we get  $$ 2=28-\left( 2 \right) { t }_{ 2 }\quad \Rightarrow { t }_{ 2 }=13\ s$$
    Total time $$ ={ t }_{ 1 }+{ t }_{ 2 }=3+13=16\ s$$ 
  • Question 7
    1 / -0
    A particle is thrown vertically upwards from ground. It takes time $${ t }_{ 1 }$$ to reach a height $$h$$. It continues to move and takes time $${ t }_{ 2 }$$ to reach the ground. Its maximum height is

    Solution
    Let $$u $$ & $$ H $$ be initial velocity and maximum height respectively so, $$ u=\sqrt { 2gH } $$
    Now, for both time $$ { t }_{ 1 }\ \&\ { t }_{ 2 }$$ displacement is  $$ h$$

    $$ h=ut-\dfrac { 1 }{ 2 } g{ t }^{ 2 }\quad $$

    $$t^2+\dfrac{2u}{g}t+\dfrac{2h}{g}=0$$ 
    which is having two roots $$t_1$$and $$t_2$$
    $${ t }_{ 1 }+{ t }_{ 2 }=\dfrac { 2u }{ g } \quad or\quad { t }_{ 1 }+{ t }_{ 2 }=\dfrac { 2\sqrt { 2gH }  }{ g } \Rightarrow H=\dfrac { g \left( { t }_{ 1 }+{ t }_{ 2 } \right)^{2} }{ 8 } $$
  • Question 8
    1 / -0
    A player throws a ball upwards with an initial speed of $$29.4\ m{ s }^{ -1 }$$. The height to which the ball rises and the time taken to reach the player's hands are
    Solution
    Maximum height $$ h=\dfrac { { u }^{ 2 } }{ g } =\dfrac { { \left( 29.4 \right)  }^{ 2 } }{ 2\left( 9.8 \right)  } =44.1m$$
    Time taken $$ t=2\sqrt { \dfrac { 2h }{ g }  } =2\sqrt { \dfrac { 2\left( 44.1 \right)  }{ 9.8 }  } =6s$$
  • Question 9
    1 / -0
    A particle experiences a fixed acceleration for $$6$$ seconds after starting from rest. It cover a distance of $${ s }_{ 1 }$$ in first two seconds, $${ s }_{ 2 }$$ in the next 2 seconds and $${ s }_{ 3 }$$ in the last 2 seconds then $${ s }_{ 3 }:{ s }_{ 2 }:{ s }_{ 1 }$$ is
    Solution
    Using the equations 
    $$s=ut+\frac { 1 }{ 2 } a{ t }^{ 2 }\\ v=u+at$$

    We get
    For $$t=0$$ to $$t=2$$
    $$u=0\quad m/s\\ s=2a\quad m$$

    For $$t=2$$ to $$t=4$$
    $$u=2a\quad m/s\\ s=6a\quad m$$

    For $$t=4$$ to $$t=6$$
    $$u=4a\quad m/s\\ s=10a\quad m$$
  • Question 10
    1 / -0
    A body of mass $$3kg$$ falls from the multi-storeyed building $$100m$$ high and buries itself $$2m$$ deep in the sand. The time of penetration is
    Solution
    $$V=\sqrt { 2gH } =\sqrt { 2000 } m/s\\ { v }^{ 2 }-{ u }^{ 2 }=2as\\ \Rightarrow 0-2000=2\times a\times 2\\ \Rightarrow a=-500\quad m/{ s }^{ 2 }\\ v=u+at\\ \Rightarrow t=\dfrac { \sqrt { 2000 }  }{ 500 } =0.089s \sim  0.09s$$
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