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An object moving with a speed of 6.25m/s, is decelerated at a rate given by dv/dt =- 2.5√v, where v is the instantaneous speed. The time taken by the object, to come to rest, would be

  • 2 s

  • 4 s

  • 8 s

  • 8 s


A.

2 s

dv over dt space equals space minus space 2.5 square root of straight v
fraction numerator dv over denominator square root of straight v end fraction space equals space minus 2.5 dt
integral subscript 6.25 end subscript superscript 0 straight v to the power of negative 1 divided by 2 end exponent space dv space equals space minus 2.5 integral subscript 0 superscript straight t dt
minus 2.5 left square bracket straight t right square bracket subscript 0 superscript straight t space equals space left square bracket 2 space straight v to the power of 1 divided by 2 end exponent right square bracket subscript 6.25 end subscript superscript 0
straight t space equals 2 straight s
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A particle of mass m is at rest at the origin at time t = 0. It is subjected to a force F (t) = F0e–bt in the x direction. Its speed v(t) is depicted by which of the following curves?


C.



So, velocity increases continuously and attains a maximum value of v= F/mb as t ∞
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A particle is moving with velocity,straight v space equals space straight k left parenthesis straight y space bold i with hat on top space plus space straight x space bold j with hat on top right parenthesis where K is a constant. The general equation for its path is

  • y = x2 + constant

  • y2 =  x + constant

  • xy = constant

  • xy = constant


D.

xy = constant

The velocity of the particle,straight v space equals space straight k left parenthesis straight y space bold i with hat on top space plus space straight x space bold j with hat on top right parenthesis 
rightwards double arrow space dx over dt space equals space ky comma space dy over dt space equals space kx
dy over dx space equals space dy over dt straight x dt over dx space equals space kx over ky
space ydy space equals space xdx
straight y squared space equals space straight x squared space space plus space straight C

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A boy can throw a stone up to a maximum height of 10 m. The maximum horizontal distance that the boy can throw the same stone up to will be

  • 20 square root of 2m
  • 10 m

  • 10 square root of 2 m
  • 10 square root of 2 m

D.

10 square root of 2 m

Maximum speed with which the boy can throw stone is

Range is maximum when projectile is thrown at angle of 45o, Thus,

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The figure shows the position –time (x – t) graph of one–dimensional motion of a body of mass 0.4 kg. The magnitude of each impulse is

  • 0.4 (Ns)

  • 0.8 Ns

  • 1.6 Ns

  • 1.6 Ns


B.

0.8 Ns

From the graph it is a straight line, so uniform motion. Because of impulse direction of velocity changes as can be seen from the slope of the graph.
(∴ impulse (l) = f x t
⇒ I  = mat
 space equals space fraction numerator straight m space left parenthesis straight v subscript 2 minus straight v subscript 1 right parenthesis over denominator straight t end fraction space straight t space equals space mv subscript 2 space minus space mv subscript 1
Initial velocity, v1 = 2/2 = 1 ms-1
Final velocity v2 = 2/2 = - 1 ms-1
pi  = mv1 = 0.4 N-s
pf  = mv2 = -0.4 N-s
J = pf-pi = -0.4-0.4
 = - 0.8 N-s
|J| = 0.8 N-s

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