Question

At t= 0 a particle leaves the origin with a velocity of 6 m/s in the positive y direction. Its acceleration is given by a = 2 – 3 m/s2. The x and
y coordinates of the particle at the instant the particle reaches maximum y coordinate are​

Answer 1

Answer:

Att=0 a particle leaves the origin with a velocity of 6 m/s in the positive y direction. Its acceleration is given by a = 2-3jm/s.

Answer 2

When the particle reaches the maximum y coordinate the coordinate of (x, y) will be (4, 6).

Given:

At t=0, the velocity u = 6j m/s

Acceleration, a= 2i-3j m/s²

To Find:

The x and y coordinates of the particle at the instant the particle reaches the maximum y coordinate.

Solution:

We know, $s= ut+\frac{1}{2} at^2$

Let us assume after time 't' the particle reaches the maximum y coordinate.

$s_x=0+\frac{1}{2} 2t^2$

$s_y= 6t+ \frac{1}{2} (-3)t^2$

$v_y= \frac{dy}{dt} = 6- 3t$

When the particle reaches the maximum y coordinate the velocity will be 0.

Therefore,

∴$v_y = 6-3t=0\\t=2$

After 2 sec the particle reaches the maximum y coordinate.

After 2 sec,

$s_x= \frac{1}{2}*2*2^2= 4\\s_y=(6*2)- \frac{1}{2}*3*2^2= 12-6= 6$

Hence, When the particle reaches the maximum y coordinate the coordinate of (x, y) will be (4, 6).

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