Home
Class 12
PHYSICS
An object has a velocity, v = (2hati + 4...

An object has a velocity, `v = (2hati + 4hatj) ms^(-1)` at time `t = 0s`. It undergoes a constant acceleration `a = (hati - 3hatj)ms^(-2)` for 4s. Then
(i) Find the coordinates of the object if it is at origin at `t = 0`
(ii) Find the magnitude of its velocity at the end of 4s.

Text Solution

Verified by Experts

(i) Here original position of the object
`bar(r)_(0)=x_(0)hat(i)+y_(0)hatj=0hati+0hat(j)`
Initial velocity `barv_(0)=v_(0x)hati+v_(0_x)hatj=2hati+4hatj`
`bara=a_(x)hati+a_(y)hatj=hati-3hatj`
And t=4 s
Let the final co -ordinates of the object be (x,y) . Then according to the equation (iii) derived in previous section.
`x = x_(0)+v_(0_(x))t+(1)/(2)a_(x)t^(2)=0+2xx4+(1)/(2)(1) xx4^(2) `
x=16
and `y=y_(0)+v_(0_y)t+(1)/(2)a_(y)t^(2)=0+4xx4+(1)/(2)(-3)xx4^(2)`
`y = -8`
Therefore the object lies at (16,-8) at t= 4 s.
(ii) Using equation ,
`bar(v)= bar(v_(0))+bar(at)`
`implies barv=(2hati+4hatj)+(hati-3hatj)xx4`
`= (2hatj+4hatj)+(4hati-12)`
`= (2+4)hati+(4-12 )hatj`
`implies barv= 6hati-8hatj:` Velocity at the end of 4 s.
`:. |barv|=sqrt(6^(2)+8^(2)) = 10 m.s^(-1)`
Its direction with x-axis `theta = tan^(-1) ""(-8)/(6)`
Promotional Banner

Similar Questions

Explore conceptually related problems

A particle starts from the origin at t = 0 s with a velocity of 10.0 hatj m/s and moves in plane with a constant acceleration of (8hati + 2hatj)ms^(-2) . The y-coordinate of the particle in 2 sec is

A particle's velocity changes from (2hat I +3 hatj) ms^(-1) in to (3 hati - 2hatj) ms^(-1) in 2s. Its average acceleration in ms^(-2) is

The velocity of an object at t =0 is vecv_(0) =- 4 hatj m/s . It moves in plane with constant acceleration veca = ( 3hati + 8 hatj) m//s^(2) . What is its velocity after 1 s?

A particle velocity changes from (2 hati - 3hatj) ms^(-1) to (3hati - 2hatj) ms^(-1) in 2s. If its mass is 1kg, the acceleraton (ms^(-2)) is

A particle starts from the origin at t=Os with a velocity of 10.0 hatj m//s and moves in the xy -plane with a constant acceleration of (8hati+2hatj)m//s^(-2) . What time is the x -coordinate of the particle 16m ?

A particle starts from origin at t = 0 with a velocity of 15 hati ms^(-1) and moves in xy-plane under the action of a force which produces a constant acceleration of 15 hati + 20 hatj ms^(-2) . Find the y-coordinate of the particle at the instant its x-coordinate is 180 m.

Velocity of a particle at time t=0 is 2 hati m/s. A constant acceleration of 2 m/s^2 acts on the particle for 2 s at an angle of 60^@ with its initial velocity. Find the magnitude of velocity and displacement of particle at the end of t=2s.

A particle P is at the origin starts with velocity u=(2hati-4hatj)m//s with constant acceleration (3hati+5hatj)m//s^(2) . After travelling for 2s its distance from the origin is

The position of an object is given by vecr= ( 9 thati + 4t^(3)hatj) m. find its velocity at time t= 1s.

Velocity of particle at time t=0 is 2ms^(-1) .A constant acceleration of 2ms^(-2) act on the particle for 1 second at an angle of 60^(@) with its initial velocity .Find the magnitude velocity and displacement of the particle at the end of t=1s.