Question

The y-coordinate of a particle is given by y = 6t3 – 5t. If ax = 14t m/sec2 and vx = 4 m/sec at t = 0.
Determine the velocity and acceleration of the particle when t = 1 sec.

Answer 1

Answer:

The Velocity and Acceleration of the particle at t=1s are 13.60m/s and 38.63m/s² respectively.

Explanation:

Given $v_{x}$=4m/s and $a_{x}$=14tm/s²

At t=1s,

$v_{x}$=4m/s and $a_{x}$=14m/s².

Also we have,

y=6t³-5t

$v_{y}$=$\frac{d}{dt}$(y)=$\frac{d}{dt}$(6t³-5t)

⇒$v_{y}$=18t²-5m/s

$a_{y}$=$\frac{d}{dt}$($v_{y}$)=$\frac{d}{dt}$(18t²-5)

⇒$a_{y}$=36tm/s²

At t=1s

$v_{y}$=18×1²-5=13m/s  and $a_{y}$=36×1=36m/s²

Velocity of the particle=$\sqrt{v_{x}^{2}+v_{y}^{2}}$

⇒v=$\sqrt{4^{2} +13^{2} }$

∴v=13.60m/s

Acceleration of the particle=$\sqrt{a_{x}^{2} +a_{y}^{2} }$

⇒a=$\sqrt{13^{2} +36^{2} }$

∴a=38.63m/s²