A particle leaves the origin with an initial velocity →v = ( 3.00^i ) m/s and a constant acceleration →a = ( −1.00^i−0.500^j ) m/s2. When it reaches its maximum x coordinate, match its position vector and velocity vector(Coloumn A) with vectors (Coloumn B)?
Coloumn A | Coloumn B |
(i) Position vector | (U) −10^j |
(ii) Velocity vector | (V) 4.5^i−22.5^j |
(W) 4.5^i−21.5^j | |
(X) −15^j |
(i) - V, (ii) - X
→u = 3^i m/s , →a = ( −1^i−5^j ) m/s2
This can be treated as 2 motions - one in x direction and the other in y direction.
x direction
sx=ut+12at2 ( Let sx be the displacement in x direction )
u = 3m/s , a = -1m/s2 ( x component of inital velocity is 3m/s and x component ot →a is -1m/s2 )
∴ sx=3t+12(−1)t2
⇒ sx=3t−12t2
vx=u+at
= 3 + (-1)t
=(3-t)s
x component will be maximum , when dsxdt or vx is zero
If we draw a graph between sx and t , we get
We can see that sx is maximum at t = 3 or when dsxdt = 0 or vx = 0
Similarly in y direction
u = 0 ( as y component of inital velocity is zero )
a = -5m/s2 ( y component of acceleration )
sy=ut+12at2
= 0 * (t) + 12(−5)t2
= −52t2m
And vy or dsydt = u + at
= 0 + (-5)t
= -5t m/s
Now at t = 3s
sx = 3t- 12t2 = 3(3) - 12 (3)^2 = 9 - 92 = 4.5m
sy = −52(t)2=−52×9=−453=−22.5m
vx = 3 - t = 3 - 3 = 0 m/s
vy = -5t = -5(3) = -15m/s
∴ Position vector = (4.5^i - 22.5^j)m
Velocity vector = -15^j m/s