A particle moves along a straight line OX. At a time t (in seconds) the distance x (in metres) of the particle from O is given by
        straight x equals 40 straight t space plus 12 straight t minus straight t cubed
How long would the particle travel before coming to rest?

  • 24m

  • 40m

  • 56m

  • 56m


C.

56m

Velocity is rate of change of distance or displacement.
Distance travelled by the particle is
                      
We know that, velocity is rate of change of distance i.e., 


but final velocity v = 0

Hence, distance travelled by the particle before coming to rest is given by
                   

            

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The x and y coordinates of the particle at any time are x = 5t – 2t2 and y = 10t respectively, where x and y are in meters and t in seconds. The acceleration of the particle at t = 2 s is

  • 0

  • 5 m/s2

  • -4 m/s2

  • -4 m/s2


C.

-4 m/s2

x = 5t – 2t2    y = 10t



Acceleration of particle at t = 2 s is = –4 m/s2

1927 Views

A body of mass 3 kg is under a constant force which causes a displacement s in metres in it, given by the relation straight s equals 1 third straight t squared, where t is in s. Work done by the force in 2 s is:

  • 5 over 19 straight J
  • 3 over 8 straight J
  • 8 over 3 straight J
  • 8 over 3 straight J

C.

8 over 3 straight J

If a constant force is applied on the object causing a displacement in it, then it is said that work has been done on the body to displace it.
  Work done by the force  = Force x Displacement
or          W = F x s                            ...(i)
But from Newton's 2nd law, we have
  Force  =  Mass x Acceleration
i.e.,       F = ma                                ...(ii)
Hence, from Eqs. (i) and (ii), we get

Hence, Eq (ii) becomes
                
                   
We have given
                   
 

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A toy car with charge q moves on a frictionless horizontal plane surface under the influence of a uniform electric field E . Due to the force q E, its velocity increases from 0 to 6 m/s in one-second duration. At that instant, the direction of the field is reversed. The car continues to move for two more seconds under the influence of this field. The average velocity and the average speed of the toy car between 0 to 3 seconds are respectively

  • 2 m/s, 4 m/s

  • 1 m/s, 3 m/s

  • 1.5 m/s, 3 m/s

  • 1 m/s, 3.5 m/s


B.

1 m/s, 3 m/s

v = - 6ms-1

Acceleration, 

a = v-ut = 6-01 = 6 ms-2For t = 0 to t = 1sS1 = 12 x 6 (1)2 = 3m ... (i)For t = 1s = to  t = 2s,S2 = 6.1 -12 x 6 (1)2 = 3m .... (ii)For t = 2s to t = 3sS3 = 0 - 12 x 6 (1)2 = - 3m (iii)Total displacementS = S1 + S2 + S3  = 3m vaverage = 3/3 = 1 ms-1Total  distance travelled = 9 mAverage  speed  = 9/3 =3 ms-1


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A ball is dropped from  a high-rise platform at t=0 starting from rest. After 6s another ball is thrown downwards from the same platform with a speed v. The two balls meet at t= 18s. What is the value of v? (take g= 10 ms-2)

  • 74 ms-2

  • 55 ms-1

  • 40 ms-1

  • 40 ms-1


A.

74 ms-2

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