An object's velocity is given by #v(t) = (2t^2 +t +1 , sin2t )#. What is the object's rate and direction of acceleration at #t=6 #?

Question

An object's velocity is given by #v(t) = (2t^2 +t +1 ,    sin2t )#.  What is the object's rate and direction of acceleration at #t=6  #?

Charles Aeschliman · Accepted Answer

#a = 27.1# #"LT"^-2#

#theta = 3.58^"o"# We're asked to find the the magnitude (rate) and direction of an object's acceleration at a given time, given its velocity equation. To do this, we need too differentiate the velocity component equations, which are given as #v_x(t) = 2t^2 + t + 1# #v_y(t) = sin(2t)# Finding the derivatives: #a_x(t) = d/(dx) [2t^2 + t + 1] = 4t + 1# #a_y(t) = d/(dx)[sin(2t)] = 2cos(2t)# (I'll assume this is in radians) Plugging in #t = 6# (no units), we have #a_x = 4(6) + 1 = 27# #"LT"^-2# #a_y = 2cos(2(6)) = 1.69# #"LT"^-2# (The #"LT"^-2# term is the dimensional form of the units for acceleration (#"distance"xx"time"^-2#). I used this term here since no units were given.) The magnitude of the acceleration is thus #a = sqrt((a_x)^2 + (a_y)^2) = sqrt(27^2 + 1.69^2) = color(red)(27.1# #color(red)("LT"^-2# And the direction is #theta = arctan((a_y)/(a_x)) = arctan(1.69/27) = color(blue)(3.58^"o"# Always make sure your arctangent calculation is in the right direction; it could be #180^"o"# off!

Xavier Adame · Answer

undefined The object's acceleration is (4, -4). Its rate of acceleration is 4 units per second squared, and its direction is opposite to the direction of the velocity vector.

Axel Acevedo · Answer

undefined To find the object's rate and direction of acceleration at $ t = 6 $, we need to calculate the second derivative of the position function $ v(t) $ with respect to time. The second derivative gives us the acceleration vector.

1. First, find the first derivative of $ v(t) $ with respect to time $ t $ for both components:

$$ v(t) = (2t^2 + t + 1, \sin(2t)) $$

$$ v'(t) = (4t + 1, 2\cos(2t)) $$

2. Then, find the second derivative of $ v(t) $ with respect to time $ t $ for both components:

$$ v''(t) = (4, -4\sin(2t)) $$

3. Evaluate the acceleration vector $ v''(6) $ at $ t = 6 $:

$$ v''(6) = (4, -4\sin(12)) $$

4. Calculate the magnitude of acceleration:
$$ |v''(6)| = \sqrt{4^2 + (-4\sin(12))^2} $$

5. Determine the direction of acceleration:
$$ \text{Direction of acceleration} = \tan^{-1}\left(\frac{-4\sin(12)}{4}\right) $$

6. Calculate the numerical values:
$$ |v''(6)| \approx \sqrt{16 + 16\sin^2(12)} $$
$$ |v''(6)| \approx \sqrt{16 + 16(0.2108)} $$
$$ |v''(6)| \approx \sqrt{16 + 3.369} $$
$$ |v''(6)| \approx \sqrt{19.369} $$
$$ |v''(6)| \approx 4.4 \, \text{(approximately)} $$

$$ \text{Direction of acceleration} \approx \tan^{-1}\left(\frac{-4\sin(12)}{4}\right) $$
$$ \text{Direction of acceleration} \approx \tan^{-1}(-0.688) $$
$$ \text{Direction of acceleration} \approx -35.05^\circ $$

Therefore, at $ t = 6 $, the object's rate of acceleration is approximately $ 4.4 $ and its direction of acceleration is approximately $ -35.05^\circ $.