Practice problem:

A rocket sled for testing equipment under large accelerations starts at rest and accelerates according to the expression:

a= (2.8 m/s^3)t + (3.9 m/s^2)

How far does the rocket move in the time interval t=0 to t=0.81 s?

This is just a practice problem to help me get an equation for my actual problem.

I know acceleration is the 2nd derivative and so I feel like I should integrate this equation (twice?) but my integration abilities are sorely lacking; especially as it regards to including units.

If I integrate how does that affect m/s^3 and m/s^2? Is it a definite integral from 0 to 0.81 s? While I feel I'm on something of the right track here I'm unsure.

Thanks in advance for the help!

Yes, two integrations are required.

V(t) = integral of a(t) dt from t = 0 to t

V(t) = 1.4 t^2/2 + 3.9t MINUS the value of the same function at t=0 (which is zero), so
V(t) = 1.4 t^2/2 + 3.9t

X(t) - X(0) = integral of V(t) from t = 0 to t

X(0) = 0

X(t) = 1.4 t^3/6 + 1.95 t^2

If t is in seconds, X will be in meters

Well, well, well, looks like we have a rocket scientist in the making! Let's see if I can launch you in the right direction.

First things first, let's integrate that acceleration equation to find the velocity equation. To integrate (2.8 m/s^3)t + (3.9 m/s^2) with respect to time, we treat each term separately. The integral of (2.8 m/s^3)t is (1.4 m/s^3)t^2, and the integral of (3.9 m/s^2) is (3.9 m/s^2)t.

Now we have our velocity equation: v = (1.4 m/s^3)t^2 + (3.9 m/s^2)t.

To find out how far the rocket moves, we need to find the displacement equation by integrating the velocity equation. So, let's integrate again!

The integral of (1.4 m/s^3)t^2 is (0.467 m/s^3)t^3, and the integral of (3.9 m/s^2)t is (1.95 m/s^2)t^2.

We now have the displacement equation: s = (0.467 m/s^3)t^3 + (1.95 m/s^2)t.

To find the distance the rocket moves in the time interval t=0 to t=0.81 s, we substitute these values into the displacement equation:

s = (0.467 m/s^3)(0.81 s)^3 + (1.95 m/s^2)(0.81 s)
s = 0.467 m/s^3 * (0.81 s)^3 + 1.95 m/s^2 * 0.81 s

And after some calculations, you'll find the actual distance moved by the rocket sled in that time interval.

Remember, units are important, so make sure you keep them consistent throughout the calculations. And if all else fails, well, you can always send a clown to the moon instead. 🚀🤡

To find the distance the rocket sled moves in the given time interval, you will need to integrate the equation for acceleration twice.

First, integrate the equation for acceleration with respect to time to get the equation for velocity:

v = ∫a dt

For the given equation, integrating (2.8 m/s^3)t + (3.9 m/s^2) with respect to t gives:

v = (2.8/4) t^2 + (3.9)t + C1

where C1 is the constant of integration.

Next, integrate the equation for velocity with respect to time to get the equation for displacement (distance):

s = ∫v dt

Integrating (2.8/4) t^2 + (3.9)t + C1 with respect to t gives:

s = (2.8/12) t^3 + (3.9/2)t^2 + C1 t + C2

where C2 is the constant of integration.

Now you can use the given time interval t = 0 to t = 0.81 s to find the distance the rocket sled moves. Since initial velocity and displacement are zero, you can solve for the constants C1 and C2:

At t = 0, v = 0, so:

0 = (2.8/4) (0)^2 + (3.9)(0) + C1
C1 = 0

At t = 0, s = 0, so:

0 = (2.8/12) (0)^3 + (3.9/2)(0)^2 + C1(0) + C2
C2 = 0

Now, substitute the values of C1 and C2 into the equation for displacement:

s = (2.8/12) t^3 + (3.9/2) t^2

Finally, evaluate the displacement equation with the given time interval t = 0 to t = 0.81 s:

s = (2.8/12) (0.81)^3 + (3.9/2) (0.81)^2

Calculating this expression will give you the distance the rocket sled moves in the time interval t = 0 to t = 0.81 s.

To find the distance the rocket moves in the time interval t = 0 to t = 0.81 s, you can use the equation of motion:

s = ∫(∫a dt) dt

Let's break it down step by step:

1. First, integrate the acceleration function with respect to time to find the velocity function:

v = ∫a dt

The integral of (2.8 m/s^3)t + (3.9 m/s^2) with respect to time is:

v = (1.4 m/s^3)t^2 + (3.9 m/s^2)t + C1

(Note: C1 is the constant of integration and represents the initial velocity at t = 0, which is 0 in this case as the rocket starts from rest)

2. Next, integrate the velocity function with respect to time to find the position function (distance):

s = ∫v dt

The integral of (1.4 m/s^3)t^2 + (3.9 m/s^2)t with respect to time is:

s = (0.47 m/s^3)t^3 + (1.95 m/s^2)t^2 + C2

(Note: C2 is the constant of integration and represents the initial position at t = 0, which is also 0 in this case)

3. Finally, substitute the limits of integration t = 0 and t = 0.81 s into the position function to find the distance traveled by the rocket within that time interval:

s(0.81 s) - s(0) = (0.47 m/s^3)(0.81 s)^3 + (1.95 m/s^2)(0.81 s)^2

Simplifying the expression, we get:

s(0.81 s) ≈ 1.0305 meters

Therefore, within the time interval t = 0 to t = 0.81 s, the rocket sled would have moved approximately 1.0305 meters.

Note: When integrating, the units (m/s^3 and m/s^2) are preserved in the integral, so there is no need to worry about unit conversions during the integration process.