Advertisements
Advertisements
प्रश्न
A particle starts from the origin at t = 0 s with a velocity of 10.0 `hatj "m/s"` and moves in the x-y plane with a constant acceleration of `(8.0 hati + 2.0 hatj) ms^(-2)`.
- At what time is the x-coordinate of the particle 16 m? What is the y-coordinate of the particle at that time?
- What is the speed of the particle at the time?
Advertisements
उत्तर १
Velocity of the particle `vecv = 10.0 hatj` m/s
Acceleration of the particle = `veca = (8.0 hati + 2.0 hatj)`
Also
But `veca = (dvecv)/(dt) = 8.0 hati +2.0 hatj`
`(dvecv) = (8.0 hati + 2.0 hatj)dt`
Integrating both sides:
`vecv(t)= 8.0t hati + 2.0t hatj + vecu`
where
`vecu` = velocity vector of the particle at t= 0
`vecv` = velocity vector of the particle at time t
But `vecv = (dvecr)/(dt)`
`dvecr = vecvdt = (8.0t hati + 2.0t hatj + vecu)dt`
Integrating the equations with the conditions: at t = 0; r = 0 andat t = t; r = r
`vecr = vecut + 1/28.0t^2 hati + 1/2xx2.0t^2 hatj`
`=vecut + 4.0t^2 hati + t^2 hatj`
`=(10.0 hatj)t + 4.0t^2 hati + t^2 hatj`
`x hati + y hatj = 4.0t^2 hati + (10t + t^2)hatj`
Since the motion of the particle is confined to the x-y plane, on equating the coefficients of `hati "and" hatj`, we get:
`x = 4t^2`
`t = (x/4)^(1/2)`
And `y = 10t + t^2`
(a) When x = 16 m
`t=(16/4)^(1/2)= 2s`
∴ y = 10 × 2 + (2)2 = 24 m
(b) Velocity of the particle is given by:
`vecv(t) = 8.0t hati + 2.0t hatj + hatu`
at t = 2s
`vecv(t) = 8.0 xx 2 hati + 2.0 xx 2 hatj + 10 hatj`
=`16 hati+ 14 hatj`
∴Speed of the particle
`|vecv| = sqrt((16)^2 + (14)^2)`
`=sqrt(256+196) = sqrt(452)`
`= 21.26 "m/s"`
उत्तर २
it is given that `vecr_(t = 0s) = vecv_(0) = 10.0 hatj` m/s and `veca(t) = (8.0 hati + 2.0 hatj) ms^(-2)`
(a) it means `x_0 = 0,u_x = 0, a_x = 8.0` `ms^(-2)` and x = 16 m
Using relation `s = x - x_0 = u_xt+1/2a_xt^2` we have
`16 - 0 = 0 + 1/2 xx 8.0 xx t^2 => t = 2s`
`:.y = y_0 + u_yt+ 1/2a_yt^2 = 0 + 10.0xx2+1/2xx2.0xx(2)^2`
= 20 + 4 = 24 m
(b) Velocity of particle at t= 2 s along x-axis
`v_x = u_x+a_xt=0 + 8.0 xx 2 = 16.0` m/s
and along y-axis `v_y = u_y+a_yt = 10.0 + 2.0 xx 2 = 14.0` m/s
∴Speed of particle at t = 2s
`v= sqrt(v_x^2+v_y^2) = sqrt((16.0)^2+(14.0)^2) = 21.26 ms^(-1)`
APPEARS IN
संबंधित प्रश्न
Is it possible to have an accelerated motion with a constant speed? Explain
A uniform circular motion is an accelerated motion. Explain it. State whether the acceleration is uniform or variable? Name the force responsible to cause this acceleration. What is the direction of force at any instant? Draw a diagram in support of your answer.
Which of the following quantity remains constant in uniform circular motion:
Which of the following quantity remains constant in a uniform circular motion?
Which of the following remains constant in a uniform circular motion, Speed or Velocity, or both?
A particle goes round a circular path with uniform speed v. After describing half the circle, what is the change in its centripetal acceleration?
A particle is moving in uniform circular motion with speed 'V' and radius 'R'. The angular acceleration of the particle is ______.
Certain neutron stars are believed to be rotating at about 1 rev/s. If such a star has a radius of 1.6 km, the acceleration of an object on the equator of the star will be nearly ____________.
A disc has mass 'M' and radius 'R'. How much tangential force should be applied to the rim of the disc, so as to rotate with angular velocity 'ω' in time t?
A body of mass ·m' is moving along a circle of radius 'r' with linear speed 'v'. Now, to change the linear speed to `V/2` and to move it along the circle of radius '4r', required change in the centripetal force of the body is ______.
Two particles P and Q are moving in concentric circles of rarui rp and rQ respectively. If their period of revolutions are in ratio 2 : 3, then ratio of their centripetal acceleration is ____________.
A string of length `l` is fixed at one end and carries a mass 'm' at the other end. The mass is revolving along a horizontal circle of radius 'r' making 'θ' as the semi-vertical angle of cone and `(1/pi)` revolutions per second around the vertical axis through fixed end. The tension in the string is ______.
In uniform motion, an object travels equal ______ in ______ interval of time.
A body moving along a circular path of radius R with velocity v, has centripetal acceleration a. If its velocity is made equal to 2v. What will be the centripetal acceleration?
A wheel rotating at the same angular speed undergoes constant angular retardation. After the revolution, angular velocity reduces to half its initial value. It will make ______ revolution before stopping.
A particle is performing a uniform circular motion along a circle of radius R. In half the period of revolution, its displacement and distance covered are respectively.
A wheel is Subjected to uniform angular acceleration about its axis. Initially. its angular velocity is zero. In the first 2 s, it rotates through an angle θ1, in the next 2 s, it rotates through an angle θ2. The ratio of `theta_2/theta_1` is ______.
A body of mass m is moving in circle of radius r with a constant speed v. The work done by the centripetal force in moving the body over half the circumference of the circle is ______.
In uniform circular motion, although the speed is constant, why does acceleration occur?
By applying Kepler's third law to the formula for centripetal force, Newton concluded that the force acting on a planet is:
