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

Question

1776 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-07T14:40:45

$acceleration = (1, 67) = (\to i + 67 \to j) m s^{- 2}$ $"acceleration"=(1,67)=(veci+67vecj)ms^(-2)$

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

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

ie. $a (t) = . v (t) = \frac{d (v (t))}{d t}$ $" "a(t)=dotv(t)=(d(v(t)))/(dt)$

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

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

$a (7) = (1, 70 - 3) = (1, 67) = (\to i + 67 \to j) m s^{- 2}$ $a(7)=(1,70-3)=(1,67)=(veci+67vecj)ms^(-2)$

direction ${tan}^{- 1} (\frac{67}{1}) = {tan}^{- 1} 67$ $tan^(-1)(67/1)=tan^(-1)67$ with x-axis

An object's two dimensional velocity is given by v(t) = ( t-2 , 5t^2-3t)v(t)=(t−2,5t2−3t)v(t) = ( t-2 , 5t^2-3t). What is the object's rate and direction of acceleration at t=7 t=7t=7 ?