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
A cyclist is riding with a speed of 27 km/h. As he approaches a circular turn on the road of radius 80 m, he applies brakes and reduces his speed at the constant rate of 0.50 m/s every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn?
Earth moves around the sun with uniform velocity.
The motion of the moon around the earth in a circular path is an accelerated motion.
A uniform metre rule of mass 100g is balanced on a fulcrum at mark 40cm by suspending an unknown mass m at the mark 20cm.
To which side the rule will tilt if the mass m is moved to the mark 10cm ?
Which of the following remains constant in a uniform circular motion, Speed or Velocity, or both?
The ratio of the angular speed of minute hand and hour hand of a watch is ____________.
A mass 'm' is tied to one end of a spring and whirled in a horizontal circle with constant angular velocity. The elongation in the spring is 1 cm. If the angular speed is doubled, the elongation in the spring is 6 cm. The original length of the spring is ______.
A stone of mass 3 kg attached at one end of a 2m long string is whirled in horizontal circle. The string makes an angle of 45° with the vertical then the centripetal force acting on the string is ______.
(g = 10 m/s2 , tan 45° = 1)
A cyclist is riding with a speed of 43.2 km/h. As he approaches a circular turn on the road of radius 60 m, he applies brakes and reduces his speed at constant rate of 1.8 ms-2. The magnitude of the net acceleration of the cyclist 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.
It is possible to have objects moving with uniform velocity but non-uniform acceleration.
Earth can be thought of as a sphere of radius 6400 km. Any object (or a person) is performing circular motion around the axis of earth due to earth’s rotation (period 1 day). What is acceleration of object on the surface of the earth (at equator) towards its centre? what is it at latitude θ? How does these accelerations compare with g = 9.8 m/s2?
A particle is given an initial speed u inside a smooth spherical shell of radius R = 1 m that it is just able to complete the circle. Acceleration of the particle when its velocity is vertical is ______.
A wet open umbrella is held upright and is rotated about the handle at a uniform rate of 21 revolutions in 44 s. If the rim of the umbrella is a circle of 1 metre in diameter and the height of the rim above the floor is 1.5 m, the drops of water spun off the rim and hit the floor at a horizontal ______ m from the umbrella.
The distance of the Sun from earth is 1.5 × 1011 m and its angular diameter is (2000) s when observed from the earth. The diameter of the Sun will be ______.
A ceiling fan rotates about its own axis with some angular velocity. When the fan is switched off the angular velocity becomes `(1/4)^"th"` of the original in time 't' and 'n' revolutions are made in that time. The number of revolutions made by the fan during the time interval between switch off and rest are ______. (Angular retardation is uniform)
A particle moves along a circle of radius r with constant tangential acceleration. If the velocity of the particle is v at the end of second revolution, after the revolution has started, then the tangential acceleration is ______.
Angular displacement is measured in which unit?
