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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

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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

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

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RowanMaster · Tutor for 5 years

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(a) To calculate acceleration: <br />### \(a = \frac{(6i+9j) m/s}{3 s} = (2i + 3j) m/s^2\)<br />(b) To find the the trajectory coordinates at `t=3 sec', substitute \(a, u, v_f, v_i and t=3\) in the position vector.<br /><br />## Calculating x-coordinate:<br />### \(x = u_x t + \frac{1}{2} a_x t^2 = u_x3 + \frac{1}{2} a_x9 = 3i *3+ \frac{1}{2}*2i*9 = \vec{9}+\vec{9} = 18 \, m=18 i\)<br /><br />## Calculating y-coordinate:<br />### \(y = u_y t + \frac{1}{2} a_y t^2= u_y3 + \frac{1}{2} \, a_y9 = -2j *3+ \frac{1}{2} *3j*9 = -6j+13.5j=7.5=7.5j m$\)<br />Therefore, the coordinates at \(t=3 seconds\) are (18i, 7.5j).

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## Step 1: Find the acceleration.<br />The formula to find acceleration is \(a = \frac{\Delta v}{\Delta t}\), where \(Δv\) refers to the change in velocity, and \(Δt\) is the time interval. <br />To find the change in velocity, calculate \(\Delta v = v_f - v_i\) = (9i+7j)m/s - (3i-2j)a = (6i+9j)m/s. <br />We're given that the time interval, \(\Delta t\), is \(3sec). <br /><br />## Step 2: Find the coordinates.<br />The position of the object at any time \(t\) when it starts from rest is given by the equation \(s = ut + \frac{1}{2}at^2\) , but here, since the body is moving in the xy-plane, we can write the position vector as components namely: \(x,y\).<br />### The formulas for landing points are: \(x=ucos(\theta)t + \frac{1}{2} acos(\theta) t^2\) and \(y=usin(\theta) t - \frac{1}{2} gsin(\theta) t^2\).<br />Here, \(u = v_i= 3i - 2j = (3,-2)\) and \(a = (2i+3j)\) <br />From the problem statement, we know that at \(t=3 sec\), the initial position, \(x = y = 0 \).<br />The exact angle of the initial velocity can be ignored because we only need the x and y components of the velocity.
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