Velocity , Displacement problems regarding a rabbit and hawk

[mathlearn]

Here an image on what is going to happen (Sweating)

View attachment 6245

A hawk starts it's flight instantly from rest. It flew 5 seconds with uniform acceleration of 4.8 $ms^2$ and maintained its velocity for 3 seconds until it reached the prey in linear path

i. Find the velocity of the hawk in the first 5 seconds

ii. What is the displacement of the hawk when it flew in acceleration?

iii. Find the total displacement of the hawk during the 8 seconds

I don't even know how should this be started

 

[topsquark]

Here an image on what is going to happen (Sweating)
A hawk starts it's flight instantly from rest. It flew 5 seconds with uniform acceleration of 4.8 $ms^2$ and maintained its velocity for 3 seconds until it reached the prey in linear path

i. Find the velocity of the hawk in the first 5 seconds

ii. What is the displacement of the hawk when it flew in acceleration?

iii. Find the total displacement of the hawk during the 8 seconds

I don't even know how should this be started

i) The hawk starts from rest and moves with a constant acceleration in a straight line. This sound like a \(\displaystyle \Delta v = a \Delta t\) situation...

Let's see if that can get you started.

-Dan

 

[mathlearn]

i) The hawk starts from rest and moves with a constant acceleration in a straight line. This sound like a \(\displaystyle \Delta v = a \Delta t\) situation...

Let's see if that can get you started.

-Dan

Thanks :) guess you are using the formula $v=u+at$

So getting started

$v=4.8 ms^{-2} * 5 s = 24 ms^{-1}$

Need help for the remaining two questions..

ii. What is the displacement of the hawk when it flew in acceleration?

iii. Find the total displacement of the hawk during the 8 seconds

 

[HallsofIvy]

If an object moves with constant acceleration a m per s per s, with starting speed u m per s, then after t seconds, the speed is v= u+ at. The distance moved is d= ut+ (a/2)t^2. I am surprised you would know the first formula but not the second.

 

[MarkFL]

(ii) Let's look at the hawk's velocity under acceleration:

\begin{tikzpicture}[>=stealth, xscale=5, font=\large, scale=0.5]
\foreach \i in {0,1,2,3,4,5} {%
\draw (\i,.5) -- (\i,-.5) node[below] {$\i$};%
}
\foreach \i in {0,5,10,15,20,25} {%
\draw (.1,\i) -- (-.1,\i) node

{$\i$};%
}
\draw[->] (-0.5,0) -- (5.5,0) node

{$t$};
\draw[->] (0,-2.5) -- (0,27.5) node[above] {$v$};
\draw[domain=0:5, smooth, variable=\t, ultra thick, blue] plot ({\t},{4.8*(\t)}) node

{$v(t)$};
\fill [cyan, domain=0:5, variable=\t]
(0, 0)
-- plot ({\t}, {4.8*\t})
-- (5, 0)
-- cycle;
\end{tikzpicture}

The area shaded in cyan represents the displacement during this period.

(iii) Now let's look at the entire 8 seconds:

\begin{tikzpicture}[>=stealth, xscale=5, font=\large, scale=0.5]
\foreach \i in {0,1,2,3,4,5,6,7,8} {%
\draw (\i,.5) -- (\i,-.5) node[below] {$\i$};%
}
\foreach \i in {0,5,10,15,20,25} {%
\draw (.1,\i) -- (-.1,\i) node

{$\i$};%
}
\draw[->] (-0.5,0) -- (8.5,0) node

{$t$};
\draw[->] (0,-2.5) -- (0,27.5) node[above] {$v$};
\draw[domain=0:5, smooth, variable=\t, ultra thick, blue] plot ({\t},{4.8*(\t)});
\draw[domain=5:8, smooth, variable=\t, ultra thick, blue] plot ({\t},{24}) node

{$v(t)$};
\fill [cyan, domain=0:5, variable=\t]
(0, 0)
-- plot ({\t}, {4.8*\t})
-- (5, 0)
-- cycle;
\fill [cyan, domain=5:8, variable=\t]
(5, 0)
-- plot ({\t}, {24})
-- (8, 0)
-- cycle;
\end{tikzpicture}

What is the area of the trapezoid representing the hawk's total displacement?​

 

[mathlearn]

(ii) Let's look at the hawk's velocity under acceleration:

\begin{tikzpicture}[>=stealth, xscale=5, font=\large, scale=0.5]
\foreach \i in {0,1,2,3,4,5} {%
\draw (\i,.5) -- (\i,-.5) node[below] {$\i$};%
}
\foreach \i in {0,5,10,15,20,25} {%
\draw (.1,\i) -- (-.1,\i) node

{$\i$};%
}
\draw[->] (-0.5,0) -- (5.5,0) node

{$t$};
\draw[->] (0,-2.5) -- (0,27.5) node[above] {$v$};
\draw[domain=0:5, smooth, variable=\t, ultra thick, blue] plot ({\t},{4.8*(\t)}) node

{$v(t)$};
\fill [cyan, domain=0:5, variable=\t]
(0, 0)
-- plot ({\t}, {4.8*\t})
-- (5, 0)
-- cycle;
\end{tikzpicture}

The area shaded in cyan represents the displacement during this period.

(iii) Now let's look at the entire 8 seconds:

\begin{tikzpicture}[>=stealth, xscale=5, font=\large, scale=0.5]
\foreach \i in {0,1,2,3,4,5,6,7,8} {%
\draw (\i,.5) -- (\i,-.5) node[below] {$\i$};%
}
\foreach \i in {0,5,10,15,20,25} {%
\draw (.1,\i) -- (-.1,\i) node

{$\i$};%
}
\draw[->] (-0.5,0) -- (8.5,0) node

{$t$};
\draw[->] (0,-2.5) -- (0,27.5) node[above] {$v$};
\draw[domain=0:5, smooth, variable=\t, ultra thick, blue] plot ({\t},{4.8*(\t)});
\draw[domain=5:8, smooth, variable=\t, ultra thick, blue] plot ({\t},{24}) node

{$v(t)$};
\fill [cyan, domain=0:5, variable=\t]
(0, 0)
-- plot ({\t}, {4.8*\t})
-- (5, 0)
-- cycle;
\fill [cyan, domain=5:8, variable=\t]
(5, 0)
-- plot ({\t}, {24})
-- (8, 0)
-- cycle;
\end{tikzpicture}

What is the area of the trapezoid representing the hawk's total displacement?​

Thank you for the TiKZ diagrams (Yes)

How did you plot this graph ? was the coordinates

(1,4.8) (2,9.6) (3,14.4) ... respectively

Area of the trapezium = $\frac{8+3}{2}*24=(8+3)*12= 11*12=132 m $

If an object moves with constant acceleration a m per s per s, with starting speed u m per s, then after t seconds, the speed is v= u+ at. The distance moved is d= ut+ (a/2)t^2. I am surprised you would know the first formula but not the second.

Thanks for the formula

Using this formula,

Here I guess u means initial velocity which when the hawk starts from rest is equal to zero (Thinking)

$d= ut+ (\frac{at^2}{2})=-10.4*8+(\frac{4.8*8^2}{2})=153.6 m$

Looks like something went wrong In using the formula​

 

[MarkFL]

You have let the hawk accelerate for the entire 8 seconds...you want to use:

\(\displaystyle d=d_1+d_2\)

where:

$d_1$ = distance traveled under acceleration.

$d_2$ = distance traveled with no acceleration

Hence:

\(\displaystyle d=\frac{1}{2}\left(4.8\,\frac{\text{m}}{\text{s}^2}\right)\left(5\text{ s}\right)^2+\left(24\,\frac{\text{m}}{\text{s}}\right)\left(3\text{ s}\right)\)

\(\displaystyle d=(60+72)\text{ m}=132\text{ m}\)