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Question

A particle leaves the origin with an initial velocity v = ( 3.00^i ) m/s and a constant acceleration a = ( 1.00^i0.500^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^i22.5^j
(W) 4.5^i21.5^j
(X) 15^j

A

(i) - U, (ii) - V

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B

(i) - X, (ii) - W

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C

(i) - V, (ii) - X

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D

(i) - W, (ii) - X

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Solution

The correct option is C

(i) - V, (ii) - X


u = 3^i m/s , a = ( 1^i5^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=3t12t2

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


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