Finding the velocity of a rock thrown vertically on Mars

[ChiralSuperfields]

Homework Statement

Please see below

Relevant Equations

Please see below

For this problem,

The answer is ##6.28 m/s##, however, I got ##-314 m/s## and I am not sure what I have done wrong.

My working is,
##H'(t) = \lim_{t \rightarrow 1} \frac{H(t) - H(1)}{t - 1}##
##H'(t) = \lim_{t \rightarrow 1} \frac{-1.86t^2 + 10t -8.14}{t - 1}##
##H'(t) = \lim_{t \rightarrow 1} \frac{(93t - 407)(t - 1)}{t - 1}##
##H'(t) = \lim_{t \rightarrow 1} 93t - 407 = -314##

Could someone please point out what I have done wrong?

Many thanks!

 

[topsquark]

Homework Statement: Please see below
Relevant Equations: Please see below

For this problem,
View attachment 325465
The answer is ##6.28 m/s##, however, I got ##-314 m/s## and I am not sure what I have done wrong.

My working is,
##H'(t) = \lim_{t \rightarrow 1} \frac{-1.86t^2 + 10t -8.14}{t - 1}##
##H'(t) = \lim_{t \rightarrow 1} \frac{(93t - 407)(t - 1)}{t - 1}##
##H'(t) = \lim_{t \rightarrow 1} 93t - 407 = -314##

Could someone please point out what I have done wrong?

Many thanks!

Are line 1 and line 2 equal?

-Dan

 

[ChiralSuperfields]

Are line 1 and line 2 equal?

-Dan

Thank you for your reply @topsquark !

I believe the factorization is correct (I used my graphics calculator to find the roots). The derivative formula I am using is ##H'(t) = \lim_{t \rightarrow 1} \frac{H(t) - H(1)}{t - 1}##

Many thanks!

 

[topsquark]

Thank you for your reply @topsquark !

I believe the factorization is correct (I used my graphics calculator to find the roots). The derivative formula I am using is ##H'(t) = \lim_{t \rightarrow 1} \frac{H(t) - H(1)}{t - 1}##

Many thanks!

So, you are saying that
##-1.86 t^2 + 10 t - 8.41 = (93 t - 407)(t - 1)##

FOIL out that RHS, please.

-Dan

 

[ChiralSuperfields]

So, you are saying that
##-1.86 t^2 + 10 t - 8.41 = (93 t - 407)(t - 1)##

FOIL out that RHS, please.

-Dan

Thank you for your reply @topsquark!

True, they are not equal. I need to multiply the right hand ride by a factor of -1.86/93 to make the equations equal. This gives the correct solution.

Many thanks!

 

[pasmith]

Homework Statement: Please see below
Relevant Equations: Please see below

For this problem,
View attachment 325465
The answer is ##6.28 m/s##, however, I got ##-314 m/s## and I am not sure what I have done wrong.

My working is,
##H'(t) = \lim_{t \rightarrow 1} \frac{H(t) - H(1)}{t - 1}##
##H'(t) = \lim_{t \rightarrow 1} \frac{-1.86t^2 + 10t -8.14}{t - 1}##
##H'(t) = \lim_{t \rightarrow 1} \frac{(93t - 407)(t - 1)}{t - 1}##
##H'(t) = \lim_{t \rightarrow 1} 93t - 407 = -314##

Could someone please point out what I have done wrong?

Many thanks!

No need to calculate [itex]H(1)[/itex], or resort to a calculator (or the quadratic formula) to factorise the result:[tex]
\begin{split}
\lim_{t \to 1} \frac{H(t)- H(1)}{t - 1} &= \lim_{t \to 1} \frac{10t - 1.96t^2 - (10 \times 1 - 1.86 \times 1^2)}{t - 1} \\
&= \lim_{t \to 1} \frac{10(t-1) - 1.86(t^2 - 1^2)}{t-1} \\
&= \lim_{t \to 1} \frac{10(t-1) -1.86(t + 1)(t - 1)}{t-1} \\
&= \lim_{t \to 1} \left(10 - 1.86(t + 1)\right) \\
&= 10 - 2 \times 1.86.\end{split}[/tex] But you can just differentiate [itex]H[/itex] directly. [tex]\begin{split}
H'(t) &= 10 - 2 \times 1.86 t \\
H'(1) &= 10 - 2 \times 1.86.
\end{split}[/tex]

 

[FactChecker]

I am surprised that you have to go to the basic definition of the derivative. The formula for the derivative of a polynomial would be covered very early in any calculus class. You should check whether you are expected to know that formula at this point. If so, you should review basic calculus and make the theorems and formulas in it second nature to you.