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## Differential Calculus (2017 edition)

### Course: Differential Calculus (2017 edition)>Unit 11

Lesson 2: Rectilinear motion

# Motion problems: when a particle is speeding up

The position of a particle moving along the x-axis is given by s(t)=t³-6t²+9t. Sal analyzes it to find the times when the particle is "speeding up.". Created by Sal Khan.

## Want to join the conversation?

• If acceleration = s(t) is the second derivative of s(t), whats are we finding out in the third derivative or even fourth derivative of s(t) ? • The third derivative is the rate of change in acceleration. For example, if you keep the accelerator pedal in a car pressed to the floor, the car will eventually reach maximum speed and stop accelerating (or minimum speed, crumpled against a tree). Anyway, the third derivative is often (but not universally) called jerk, and the rate of change in jerk is (again, not universally) called jounce. Jerk and jounce can be important in some systems, such as controlling the flight of drones.
• Shouldn't we be speeding up between second and third second? • Think of it this way: Between second and third the particle is still moving in the left direction (because velocity is negative) but you are accelerating in the right direction (because acceleration is positive, i.e. the slope of velocity is positive). So think of it as the particle is slowing down in the left direction and therefore you are not speeding up, you are actually slowing down. It is only when the particle switches direction to the right that it is speeding up (i.e. velocity turns positive at t > 3). Hope this helps!
• At , why did he plug in 2 and 4 into the original equation to find the minimum?? Shouldn't he have just picked one number?? • Is the acceleration decreasing or increasing in the interval 1<t<2? is the acceleration=0 at t=1 and at t=2 because the slopes at these points are zero. if between t=1 n t=2 if it is becoming more and more negative how can it become zero at t=2(slope =0) because it is increasing in the negative direction? • Your question "Is the acceleration decreasing or increasing" is asking about the change in acceleration. That concept is not needed to answer the question of when is the particle "speeding up", but it is a good question. On the interval 1 < t < 2 the acceleration is negative, but it is increasing. You can see this by imagining the tangent line to the velocity function in that interval. That tangent line is slanted down, so the slope of that line is negative, so the acceleration is negative. But as you move the tangent line to the right, its slope becomes less and less negative, the slope is increasing. Acceleration is the derivative of velocity. Sal didn't do this, but you can take the derivative of the velocity function and get the acceleration function:

v'(t) = a(t) = 6t - 12

From looking at the acceleration function you can also figure out the acceleration is negative but increasing from t = 0 to t = 2. From t = 0 to 2, the acceleration is going to be negative, at t = 2 the acceleration is zero, and at t > 2 the acceleration is positive. The function for acceleration is a linear function with a slope of positive 6, so the function is always increasing. Another way to see the acceleration is always increasing is to take the derivative of the acceleration function:

a'(t) = 6

So the acceleration is always increasing at a rate of 6/1.
• Since `v(t) = ds / dt` and `a(t) = dv / dt`, could you write it as `a(t) = d(ds / dt) / dt`? • How did Sal automatically know the vertex of the parabola was at (2, -1)?
(1 vote) • So the largest exponent on the original equation points out how many times is the particle speeding up? And the coefficient on the first term shows the magnitude of the speeding? • Well, the short answer is "no and no".
The original equation describing movement on a line was
s(t) = t³ - 6t² +9t
In this case, there were two times when the particle was speeding up
That would be where the velocity and acceleration are both positive OR both negative in sign.
t(1, 2) Time between 1 and 2 seconds
t(3, ∞) Time greater than 3 seconds.
We do not count anything that happens before the clock starts ticking at time = 0

The coefficient on the first term doesn't show the magnitude of speeding. The coefficient on the first term is one. That doesn't match any way I can interpret your words, "magnitude of the speeding".

If you look at the graph of the acceleration (the rate of change for the velocity), it has a slope of 6. The only equation that has a coefficient of 6 on the first term is the second derivative of s(t), which is the equation for the acceleration, and of course its slope would equal that coefficient.
a(t) = s"(t) = 6t - 12

However, if you look at the slope of the velocity, at every point it is changing: after all, it is a parabolic function. Sal walked us through how the slope is negative and flattens out to zero and then becomes a little positive, and then very steeply positive. So if you mean from that graph that it is speeding up according to the coefficient of that equation, that also doesn't work. The equation for the velocity is v(t) = s"(t) = 3t²- 12t + 9

The main shortcut that I know is to quickly do the single and double derivatives and examine the behaviors of those curves. The powerful one for this purpose is the curve of the velocity or s'(t) as Sal showed us.
• At , why is the particle speeding up between 1<t<2? Shouldn't it be slowing down since the particle's velocity reaches 0 at t=2?
(1 vote) • How can we say that velocity is increasing only when it goes to right direction and decreases when it goes to left?  