An object's two dimensional velocity is given by $v(t) = ( t-2 , 5t^2-3t)$ . What is the object's rate and direction of acceleration at $t=7$ ?

Question

1769 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-07T14:40:45

$"acceleration"=(1,67)=(veci+67vecj)ms^(-2)$

direction $=tan^(-1)67$ with x-axis

If we are given velocity as a function of time, the acceleration is found by differentiating that function.

ie. $" "a(t)=dotv(t)=(d(v(t)))/(dt)$

$v(t)=(t-2,5t^2-3t)$

$a(t)=dotv(t)=(1,10t-3)$

$a(7)=(1,70-3)=(1,67)=(veci+67vecj)ms^(-2)$

direction $tan^(-1)(67/1)=tan^(-1)67$ with x-axis

An object's two dimensional velocity is given by v(t) = ( t-2 , 5t^2-3t)v(t) = ( t-2 , 5t^2-3t). What is the object's rate and direction of acceleration at t=7 t=7 ?