# Thread: Derivative question? Acceleration, and instantaneous and average velocity.

1. ## Derivative question? Acceleration, and instantaneous and average velocity.

At a time t seconds after it is thrown up in the air, a tomato is at a height of f(t) = -4.9t^2 + 25t + 3 meters.

(a) What is the average velocity of the tomato during the first 1.7 seconds? Include units.

I know the units are m/sec but I can't get the average velocity!

(b) Find the instantaneous velocity of the tomato at t = 1.7. Include units.

It's 8.34 m/sec.

(c) What is the acceleration at t = 1.7?

-9.8 m/sec^2

(d) How high does the tomato go?

34.89 m

(e) How long is the tomato in the air?

I can't get this!! _______ sec

I'm so lost. Thank you =)

2. Originally Posted by melliep
At a time t seconds after it is thrown up in the air, a tomato is at a height of f(t) = -4.9t^2 + 25t + 3 meters.

(a) What is the average velocity of the tomato during the first 1.7 seconds? Include units.

I know the units are m/sec but I can't get the average velocity!
Averate velocity $\displaystyle = \frac{f(1.7)-f(0)}{1.7-0}$

3. Originally Posted by melliep
At a time t seconds after it is thrown up in the air, a tomato is at a height of f(t) = -4.9t^2 + 25t + 3 meters.

(b) Find the instantaneous velocity of the tomato at t = 1.7. Include units.

It's 8.34 m/sec.
OK this is in the calculus forum and is a calculus problem, so you should know that the instantaneous velocity is $\displaystyle \frac{d}{dt}f(t)$.

Now what is the problem?

CB

4. In reference to "a", thank you, I know. But for some reason I'm not getting it correct. Could you solve it for me please?

As you said, this is in the calculus section. And is a calculus problem. I am taking calculus. Therefore, I do know a thing or two.

But the online quiz is telling me I'm incorrect and I want to see what your answer is.

Terribly sorry.

5. Originally Posted by CaptainBlack
OK this is in the calculus forum and is a calculus problem, so you should know that the instantaneous velocity is $\displaystyle \frac{d}{dt}f(t)$.

Now what is the problem?

CB
Um. I got the right answer for b. There is no problem.

6. Originally Posted by melliep
At a time t seconds after it is thrown up in the air, a tomato is at a height of f(t) = -4.9t^2 + 25t + 3 meters.

(c) What is the acceleration at t = 1.7?

-9.8 m/sec^2
OK, as befor we are studying calculus so we should know that acceleation is the second derivative of position $\displaystyle a=\frac{d^2}{dt^2}f(t)$.

You should have covered differentiation of polynomials, so what exactly is the problem you are having with this?

CB

7. Originally Posted by melliep
Thank you, I know. But for some reason I'm not getting it correct. Could you solve it for me please?
Terrible sorry.
Post what you have done and I will look at that.

CB

8. Originally Posted by melliep
At a time t seconds after it is thrown up in the air, a tomato is at a height of f(t) = -4.9t^2 + 25t + 3 meters

(d) How high does the tomato go?

34.89 m
There are several ways of doing this but as this is a calculus problem I will assume that you are to find the height when the velocity is zero. So we start by solving:

$\displaystyle \frac{d}{dt}f(t)=0$

for $\displaystyle t$ and then substitute the time that we find back into $\displaystyle f(t)$ to find the maximum height.

(e) How long is the tomato in the air?

I can't get this!! _______ sec
When the tomato lands f(t)=0. this is a quadratic one root is negative and so of no interest, the other is positive and is the time that the tomato is in the air (it is the time after it is thrown that it hits the ground)

CB

9. Originally Posted by CaptainBlack
Post what you have done and I will look at that.

CB
Captain Black, I did post what I know. I posted my answers. I just did not get 'a' or 'e'. I do not know how to do 'e' whatsoever, but for 'a' I used the formula for average velocity: f(1.7)-f(0)/1.7-0.

I tried two ways. And neither were correct.

I did (-4.9(1.7)^2 + 25(1.7) + 3 - 3) / 1.7

But that was incorrect.

Then I did -9.8(1.7) + 25 + 25 / 1.7

Also incorrect. It was probably a stupid mistake due to the fact that it's very late, but what am I doing wrong?

10. Originally Posted by melliep
Captain Black, I did post what I know. I posted my answers. I just did not get 'a' or 'e'. I do not know how to do 'e' whatsoever, but for 'a' I used the formula for average velocity: f(1.7)-f(0)/1.7-0.

I tried two ways. And neither were correct.

I did (-4.9(1.7)^2 + 25(1.7) + 3 - 3) / 1.7

But that was incorrect.
What value did you get?

CB

11. 16.67 for the first attempt.

then 19.61 for the second.

Did you do the problem? What was your result?

AND for 'e' I got 2.89?

12. Originally Posted by melliep
16.67 for the first attempt.

then 19.61 for the second.

Did you do the problem? What was your result?

AND for 'e' I got 2.89?
$\displaystyle 16.67$ m/s

and for (e) you want the positive root of:

$\displaystyle -4.9 t^2+25 t+3=0$

which WolfranAlpha tells us is $\displaystyle 5.22$ s.

(all to 2 DP)

CB

13. Thank you very much.