Kinetic Energy is defined by 1/2 times the mass times the velocity squared.
KE = (1/2)mv2
In the case of straight line, when the object is not repeating it's path then distance and displacement would be same.
Answer:
(a) a = (2i + 4.5j) m/s^2
(b) r = ro + vot + (1/2)at^2
Explanation:
(a) The acceleration of the particle is given by:

vo: initial velocity = (3.00i -2.00j) m/s
v: final velocity = (9.00i + 7.00j) m/s
t = 3s
by replacing the values of the vectors and time you obtain:
![\vec{a}=\frac{1}{3s}[(9.00-3.00)\hat{i}+(7.00-(-2.00))\hat{j}]\\\\\vec{a}=(2\hat{i}+4.5\hat{j})m/s^2](https://tex.z-dn.net/?f=%5Cvec%7Ba%7D%3D%5Cfrac%7B1%7D%7B3s%7D%5B%289.00-3.00%29%5Chat%7Bi%7D%2B%287.00-%28-2.00%29%29%5Chat%7Bj%7D%5D%5C%5C%5C%5C%5Cvec%7Ba%7D%3D%282%5Chat%7Bi%7D%2B4.5%5Chat%7Bj%7D%29m%2Fs%5E2)
(b) The position vector is given by:

where vo = (3.00i -2.00j) m/s and a = (2.00i + 4.50j)m/s^2
Answer:
Explanation: what grade are u in?