# Mechanics one

• March 15th 2009, 08:07 AM
BeckyC
Mechanics one
the question says:

calculate the magnitude and direction of the acceleration of a particle that moves so that its position vector in metres is given by

r=(8t-24^2)i + (6+4t-t^2)

where t is time in seconds.

I think i have worked out the acceleration to be -4i -2j

and so as a result is the magnitude is 4.47

but i am unsure how to work out direction. My teacher whipped through it and so i got lost... can you help please? (Happy)
• March 15th 2009, 09:38 AM
Vector acceleration
Hello BeckyC
Quote:

Originally Posted by BeckyC
the question says:

calculate the magnitude and direction of the acceleration of a particle that moves so that its position vector in metres is given by

r=(8t-24^2)i + (6+4t-t^2)j

where t is time in seconds.

I think i have worked out the acceleration to be -4i -2j

and so as a result is the magnitude is 4.47

but i am unsure how to work out direction. My teacher whipped through it and so i got lost... can you help please? (Happy)

$\vec{r} = (8t-2t^2)\vec{i} + (6 + 4t -t^2)\vec{j}$

In which case, you are right: acceleration $= \vec{\ddot{r}} = -4\vec{i} - 2\vec{j}$

and its magnitude $= |\vec{\ddot{r}}| = \sqrt{4^2 + 2^2} = \sqrt{20} \approx 4.47$

To find the direction, just think about the vector $-4\vec{i} - 2\vec{j}$: this vector is $-4$ units in the $\vec{i}$ direction (i.e. 4 units to the left) and $-2$ units in the $\vec{j}$ direction (i.e. 2 units downward.)

So, draw a right-angled triangle having one side horizontal and 4 units long, and at its left-hand end a side 2 units long vertically downwards. Join the hypotenuse: this side represents the actual acceleration vector. See the attached diagram.

So if the acceleration makes an angle $\theta$ below the horizontal, $\tan\theta = \frac{2}{4}= 0.5 \Rightarrow \theta \approx 26.6^o$.

• March 15th 2009, 09:44 AM
BeckyC
Quote:

Hello BeckyCI'm guessing from your answer so far that you have made a couple of typos in your question, and it should be

$\vec{r} = (8t-2t^2)\vec{i} + (6 + 4t -t^2)\vec{j}$

In which case, you are right: acceleration $= \vec{\ddot{r}} = -4\vec{i} - 2\vec{j}$

and its magnitude $= |\vec{\ddot{r}}| = \sqrt{4^2 + 2^2} = \sqrt{20} \approx 4.47$

To find the direction, just think about the vector $-4\vec{i} - 2\vec{j}$: this vector is $-4$ units in the $\vec{i}$ direction (i.e. 4 units to the left) and $-2$ units in the $\vec{j}$ direction (i.e. 2 units downward.)

So, draw a right-angled triangle having one side horizontal and 4 units long, and at its left-hand end a side 2 units long vertically downwards. Join the hypotenuse: this side represents the actual acceleration vector. See the attached diagram.

So if the acceleration makes an angle $\theta$ below the horizontal, $\tan\theta = \frac{2}{4}= 0.5 \Rightarrow \theta \approx 26.6^o$.