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Motion in A Straight Line Test - 39

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Motion in A Straight Line Test - 39
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  • Question 1
    1 / -0
    A body traversed half of the distance with a velocity $$V_0$$. The remaining part $$V_1$$ for half of the time and with the velocity $$V_2$$ for the other half of the time. Find the average velocity of the body over the whole journey.
    Solution
    Let $$s$$ be total distance to be traveled
    $$t_0$$ be the time for which it travels  with speed $$v_0$$
    $$t_1/2$$ be the time for which it travels with $$v_1 $$and $$v_2$$
    using distance=speed $$\times$$time :
    $$\frac{s}{2}=v_1\frac{t_1}{2}+v_2\frac{t_1}{2}$$
    $$t_1=\frac{s}{v_1 +v_2} \ldots(1)$$
    Average speed =Total distance traveled /Total time
    $$ <v>=\frac{s}{t_0 +t_1}$$
    $$<v>=\frac{s}{\frac{s}{2v_0} +\frac{s}{2(v_1 +v_2)}}$$
    $$<v>=\frac{2v_0 (v_1+v_1)}{2v_0 +v_1+ v_2}$$

  • Question 2
    1 / -0
    A car starts from rest and travels with uniform acceleration "$$\alpha$$" for some time and then with uniform retardation "$$\beta$$" and comes to rest. The time of motion is "$$t$$". Find the maximum velocity attained by it.
    Solution
    let it accelerate for time $$t_1$$ and retard for time $$t_2$$
    $$t_1+t_2= t\ldots(1)$$
    using $$v=u+at$$
    initially $$u=0$$,  $$v_{final}=\alpha t_1$$
    finally $$v=0 , u=\alpha t_1$$
    $$0=\alpha t_1 -\beta t_2$$
    $$\frac{t_2}{\alpha} =\frac{t_1}{\beta}\ldots(2)$$
    using $$(1)\&(2)$$
    $$t_1=\frac{\beta t}{\alpha +\beta}$$
    maximum velocity is :$$\alpha t_1$$
    $$v_{max}=\frac{\alpha \beta t}{\alpha + \beta}$$





  • Question 3
    1 / -0
    A passenger in moving train tosses a coin which falls behind him. It means that the motion of the train is
    Solution
     If the coin falls behind the passenger that means the train is accelerated. When the coin is tossed it has same velocity as that of train but during the time it is in air its velocity becomes less than that of train (because the train is accelerated), so it falls behind the passenger.
  • Question 4
    1 / -0
    A car moving at a certain speed stops on applying brakes within 16 m. If the speed of the car is doubled, maintaining the same retardation. then at what distance does it stop? Also, calculate the percentage change in this distance.(in percent)
    Solution
    Let $$u$$ be the speed of the car,
    After brake velocity $$=0$$ on moving distance of $$16m$$
    $$v^2=u^2-2as$$
    $$\implies u^2=2as$$
    $$\implies a=\cfrac{u^2}{2s}$$
    $$\implies a=\cfrac{u^2}{32}$$
    For some retardation and speed is doubled
    $$u'\implies 2u$$
    $$a=\cfrac{4u^2}{3s}$$
    $$1=\cfrac{4^2}{32}\times \cfrac{s}{242}$$
    $$s=64m$$
    $$\%$$ change in distance$$=\cfrac{64-16}{16}\times 100=300\%$$
  • Question 5
    1 / -0
    equation for uniform accelerated motion for the displacement covered in its nth second of its motion is?
    Solution
    Using equation of motion
    $$s=ut+\dfrac{1}{2}at^2$$
    The distance traveled in n second
    $$s_{1}=un+\dfrac{1}{2}an^2$$
    The distance traveled in (n-1) second
    $$s_{1}=u(n-1)+\dfrac{1}{2}a(n-1)^2$$
    Now, the distance traveled in $$n^{th}$$ second is
    $$s_{n}=s_{1}-s_{2}$$
    $$s_{n}=un+\dfrac{1}{2}an^{2}-(u(n-1))+\dfrac{1}{2}a(n-1)^2)$$
    $$s_{n}=u+a(n-\dfrac{1}{2})$$
    The equation for uniform accelerated motion for the displacement covered in its nth second of its motion is $$s_{n}=u+a(n-\dfrac{1}{2})$$.
    Hence, The correct option is B



  • Question 6
    1 / -0
    A stone falls freely from rest, and the total distance covered by it in the last second of its motion, equals the distance covered by it in the first three seconds of its motion. Find the time for which stone remains in the air and the total height from where stone is dropped.
    Solution
    $$S_n=$$ distance coveredin first $$3s$$ 
    $$S_n=u\times t+\cfrac{1}{2}\times 10\times 3^2$$
    Freely ball  so, $$u=0$$
    $$S_n=\cfrac{1}{2}\times 10\times= 9\\ =45m$$
    $$S_n=4+\cfrac{9}{2}(2t-1)=45$$
    $$t=5s$$ in air
    $$h=\cfrac{1}{2}gt^2=4.9\times 25=122.5m$$
  • Question 7
    1 / -0
    A body is projected vertically upwards. If $$a$$ and $$b$$ be the times at which it is at height $$h$$ above the point of projection while ascending and descending respectively, then $$h$$ is :
    (Take $$g=10 \ m/s^2$$)
    Solution
    If the body is projected vertically upwards with a velocity $$u$$ , 
    $$h=ut-\dfrac{1}{2}gt^2$$   (using formula $$S=ut+\frac{1}{2}at^2$$)
    or $$t^2-(2u/g)t+2h/g=0$$
    This is a quadratic equation of $$t$$ and it has two roots $$t_1=a$$ and $$t=b$$ (say)
    We know that, product of roots $$=$$ constant term/coefficient of $$t^2$$
    So, $$ab=(2h/g)/1$$ or $$h=gab/2$$ 
  • Question 8
    1 / -0
    A body thrown vertically up from the ground passes the height $$10.2m$$ twice at an interval of $$10s$$. What was its initial velocity? (in m/s)
    Solution
    if the body moving vertically passes twice in 10 sec then it is clear that it will go up for 5 sec and next 5 sec will be consumed by body to come down back.
    distance traveled in 5 sec starting from rest and with acceleration 10(returning from top using $$S=ut+\frac{1}{2}at^2$$), is 125 meters. so total height is $$(125+10.2=135.2)$$
    initial velocity is calculated by using $$v^2-u^2=2aS=>v=\sqrt{2704}=52m/s.$$
  • Question 9
    1 / -0
    Person $$A$$ walking along a road at $$3ms^{-1}$$ sees another person $$B$$ walking on another road at right angle to his road. Velocity of B is $$4ms^{-1}$$ when he is $$10m$$ off. They are nearest to each other when person A has covered a distance of
    Solution
    when $$A$$ is at origin with velocity $$3m/s$$ towards positive X.
    and $$B$$ has velocity $$4m/s$$ in positive Y direction and initial position $$(0,-10)$$.
    so position at any time $$t$$  for body $$A$$ is given by $$(3t,0)$$ and for body $$B$$ given by $$(0,-10+4t)$$
    distance between them at any time $$t$$ is given by $$S=\sqrt{(3t)^2+(-10+4t)^2}$$
    minimum value of $$S $$ Will when $$S^2$$ is minimum,
    $$S^2={(3t)^2+(-10+4t)^2}$$
     so minimizing $$S^2$$ we get $$\frac{dS^2}{dt}=6t+8(-10+4t)=0$$
    $$=>t=\frac{40}{19}$$
    also $$\frac{d(S^2)^2}{dt^2}=+ve$$ so this value is minimum.
    at this $$t$$, $$S=3.6m$$
    so the answer is option A.

  • Question 10
    1 / -0
    A balloon is rising with a constant acceleration of $$2m/s^{2}$$. At a certain instant when the balloon was moving with a velocity of $$4m/s$$, a stone was dropped from it in a region where $$g = 10m/s^{2}$$. The velocity and acceleration of stone as it comes out from the balloon are respectively (in m/s and $$m/s^{2}$$)
    Solution
    When the stone was in the balloon, the stone had the same velocity as that of balloon. So when the stone came out of the balloon, the velocity of stone was $$4 \ m/s$$ upwards.
    After coming out of the balloon, the stone was in free fall motion, so the stone experienced an acceleration due to gravity equal to $$g = 10m/s^2$$ in downward direction.
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