# An object is at rest at (8 ,2 ,9 ) and constantly accelerates at a rate of 3 m/s as it moves to point B. If point B is at (6 ,7 ,5 ), how long will it take for the object to reach point B? Assume that all coordinates are in meters.

Jan 15, 2018

The time is $= 2.11 s$

#### Explanation:

The distance $A B$ is

$A B = \sqrt{{\left(6 - 8\right)}^{2} + {\left(7 - 2\right)}^{2} + {\left(5 - 9\right)}^{2}} = \sqrt{4 + 25 + 16} = \sqrt{45} m$

Apply the equation of motion

$s = u t + \frac{1}{2} a {t}^{2}$

The initial velocity is $u = 0 m {s}^{-} 1$

The acceleration is $a = 3 m {s}^{-} 2$

Therefore,

$\sqrt{45} = 0 + \frac{1}{2} \cdot 3 \cdot {t}^{2}$

${t}^{2} = \frac{2}{3} \cdot \sqrt{45}$

$t = \sqrt{\frac{2}{3} \cdot \sqrt{45}} = 2.11 s$

Jan 15, 2018

Find the shortest distance between point A and B, then using calculus, use the definition of acceleration and velocity to solve for the displacement in terms of time from the given acceleration, then solve for time using the previously calculated shortest distance, to get

$t = \sqrt{2} \sqrt[4]{5}$ seconds $\approx 2.11474252688113$ seconds

#### Explanation:

Rephrasing the question, we have an object at point A, $\left(8 , 2 , 9\right)$ initially not moving, then all of a sudden gets a constant acceleration of speed of $3 \frac{m}{s} ^ 2$, moving towards point B, $\left(6 , 7 , 5\right)$. Notice that I said speed instead of velocity because we don't exactly know the direction yet. What we're told to do is calculate the amount of time required to get from point A to point B.

But first, let's talk about the initial displacement. We can see that, from point A to B, the object has moved:

$\Delta x = {B}_{x} - {A}_{x} = 6 \text{m" - 8 "m" = -2 "m}$

$\Delta y = {B}_{y} - {A}_{y} = 7 \text{m" - 2 "m" = 5 "m}$

$\Delta z = {B}_{z} - {A}_{z} = 5 \text{m" - 9 "m" = -4 "m}$

Now, we just need the time, but we may have to play around with acceleration and velocity for that. We don't yet know the velocity, and we can take the integral of acceleration to get that...

... But wait! The $3 \frac{m}{s} ^ 2$ we have is specifically pointing at the direction ("vector") that goes from point A to B. To get the (shortest) distance between the two points, we could use the Pythagorean theorem:

${s}^{2} = {\left(\Delta x\right)}^{2} + {\left(\Delta y\right)}^{2} + {\left(\Delta z\right)}^{2}$

It may feel weird that I already have it set up for all three variables even though the Pythagorean theorem is originally for two variables. However, it really is just a result of attempting to apply the Pythagorean theorem twice:

${\text{Distance}}_{x , y}^{2} = {\left(\Delta x\right)}^{2} + {\left(\Delta y\right)}^{2}$

${s}^{2} = {\text{Distance}}_{x , y}^{2} + {\left(\Delta z\right)}^{2}$

${s}^{2} = {\left(\Delta x\right)}^{2} + {\left(\Delta y\right)}^{2} + {\left(\Delta z\right)}^{2}$

Anyways, we could now substitute in for $\Delta x$, $\Delta y$ and $\Delta z$ to solve for $s$:

${s}^{2} = {\left(- 2 \text{m")^2 + (5 "m")^2 + (-4 "m}\right)}^{2}$

${s}^{2} = 4 {\text{m"^2 + 25 "m"^2 + 16 "m}}^{2}$

${s}^{2} = 45 {\text{m}}^{2}$

s = sqrt(45 "m"^2) = 3sqrt(5) "m"

Ah, so the acceleration was $3 \frac{m}{s} ^ 2$, and the distance between point A and B is $3 \sqrt{5} \text{m}$. What we need to look for, now, is the time it takes for that motion. Well, we can work backwards: pretend to be solving for displacement in terms of time, then equating that to the displacement we already have, to solve for $t$.

So, using the definition of acceleration:

$a = \frac{\mathrm{dv}}{\mathrm{dt}} \rightarrow \int \left(a\right) \mathrm{dt} = \int \frac{\mathrm{dv}}{\mathrm{dt}} \mathrm{dt} \rightarrow \int \left(a\right) \mathrm{dt} = v$

We know that $a = 3 \frac{m}{s} ^ 2$, so plug that in:

$v = \int \left(3 \frac{m}{s} ^ 2\right) \mathrm{dt} = 3 \frac{m}{s} ^ 2 \int \left(1\right) \mathrm{dt} = 3 t \frac{m}{s} ^ 2 + C$

Ah, the constant! Well, it's supposed to be whatever the starting velocity happens to be. Since we started with no motion at all - no displacement, no velocity, no acceleration - then this $C$ value is just zero.

$v = 3 t \frac{m}{s} ^ 2$

Then, what is the definition of velocity?

$v = \frac{\mathrm{ds}}{\mathrm{dt}} \rightarrow \int \left(v\right) \mathrm{dt} = \int \frac{\mathrm{ds}}{\mathrm{dt}} \mathrm{dt} \rightarrow \int \left(v\right) \mathrm{dt} = s$

Now, plug in for $v$ to solve for the displacement:

$s = \int \left(3 t \frac{m}{s} ^ 2\right) \mathrm{dt} = 3 \frac{m}{s} ^ 2 \int \left(t\right) \mathrm{dt} = \frac{3 {t}^{2}}{2} \frac{m}{s} ^ 2 + C$

Again, the constant is just zero:

$s = \frac{3 {t}^{2}}{2} \frac{m}{s} ^ 2$

However, we do know what the displacement is! Substitute $3 \sqrt{5} \text{m}$ in for $s$:

$3 \sqrt{5} \text{m} = \frac{3 {t}^{2}}{2} \frac{m}{s} ^ 2$

Multiply by $\frac{2}{3}$:

$3 \sqrt{5} \text{m} \cdot \frac{2}{3} = \frac{3 {t}^{2}}{2} \frac{m}{s} ^ 2 \cdot \frac{2}{3}$

$2 \sqrt{5} \text{m} = {t}^{2} \frac{m}{s} ^ 2$

"Multiply" by ${s}^{2} / m$:

$2 \sqrt{5} \text{m} \cdot {s}^{2} / m = {t}^{2} \frac{m}{s} ^ 2 \cdot {s}^{2} / m$

$2 \sqrt{5} {\text{s}}^{2} = {t}^{2}$

Interesting, we can now take the square root:

t = sqrt(2 sqrt(5) "s"^2) = sqrt(2 sqrt(5)) "s" = sqrt(2)root(4)(5) seconds

That's a werid amount of seconds, and it's certainly an irrational number... we could approximate it using a calculator:

$t = \sqrt{2} \sqrt[4]{5}$ seconds $\approx 2.11474252688113$ seconds

There we go! That's what our calculations tell us about the time it takes for an object with the given acceleration and initial state to travel through the two given points.