Question

The velocity v of a particle moving in the xy plane is given by v = (6.0t -4.0t2 )i + 8.5j, in m/s. Here v is in m/s and t (for positive time) is in s. What is the acceleration when t = 3.0 s? i-component of acceleration? -1.80×101 m/s^2

Answer 1

Answer:

Explanation:

Given

Velocity of the particle in vector form is given by

$v=\left ( 6t-4t^2\right )\hat{i}+8.5\hat{j}$

acceleration is rate of change of velocity thus acceleration is

$a=\frac{\mathrm{d} v}{\mathrm{d} t}$

$a=(6-8t)\hat{i}+0\hat{j}$

at $t=3\ s$

$a=6-8\times 3$

$a=-18\ m/s^2$