Activity: 1. At t=0s. a particle moving in the x-y plane with constant acceleration has a velocity ofoverrightarrow (v)_(i)=(3hat (i)-2hat (j))m/s, and is at the origin. Att=3s, the particle's velocity is overrightarrow (v)_(f)=(9hat (i)+7hat (j))m/s. Find (a)the acceleration of the particle (b) Its coordinates at t=3s

## Step 1: Find the acceleration.The formula to find acceleration is , where refers to the change in velocity, and is the time interval. To find the change in velocity, calculate = (9i+7j)m/s - (3i-2j)a = (6i+9j)m/s. We're given that the time interval, , is \(3sec). ## Step 2: Find the coordinates.The position of the object at any time when it starts from rest is given by the equation , but here, since the body is moving in the xy-plane, we can write the position vector as components namely: .### The formulas for landing points are: \(x=ucos(\theta)t + \frac{1}{2} acos(\theta) t^2\) and \(y=usin(\theta) t - \frac{1}{2} gsin(\theta) t^2\).Here, \(u = v_i= 3i - 2j = (3,-2)\) and \(a = (2i+3j)\) From the problem statement, we know that at , the initial position, .The exact angle of the initial velocity can be ignored because we only need the x and y components of the velocity.