Any four difference between velocity and acceleration

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

kow [346]

Before the engines fail, the rocket's altitude at time <em>t</em> is given by

$y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

and its velocity is

$v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t$

The rocket then reaches an altitude of 1150 m at time <em>t</em> such that

$1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

Solve for <em>t</em> to find this time to be

$t=11.2\,\mathrm s$

At this time, the rocket attains a velocity of

$v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}$

When it's in freefall, the rocket's altitude is given by

$y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

where $g=9.80\frac{\rm m}{\mathrm s^2}$ is the acceleration due to gravity, and its velocity is

$v_2(t)=124\dfrac{\rm m}{\rm s}-gt$

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for $y_2(t)$ to reach 0:

$1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s$

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

${v_f}^2-{v_i}^2=2a\Delta y$

where $v_f$ and $v_i$ denote final and initial velocities, respecitively, $a$ denotes acceleration, and $\Delta y$ the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means $y_2$ will contain the information we need to find the maximum height.

$-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)$

Solve for $y_{\rm max}$ and we find that the rocket reaches a maximum altitude of about 1930 m.

(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to $y_2(t)$ ) to be about 32.6 s. Plug this into $v_2(t)$ to find the velocity before it crashes:

$v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}$

That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

3 0

3 years ago

Help again please !!!

rusak2 [61]

D I think .... don’t be mad if I’m wrong

5 0

2 years ago

Which of the following is an example of changing momentum?!

ExtremeBDS [4]

Answer:

B!!!

Explanation:

5 0

2 years ago

Read 2 more answers

Think about the process of pulling down the spring and letting it go. For this event, list each form of energy associated with t

Afina-wow [57]

Answer:

gravitational 2.force of gravity

Explanation:

1.it ocours when an object is thrown into the sky. 2.iy ocurs when an object is falling or being pulled from the sky

6 0

2 years ago

A boy who is riding his bicycle, moves with an initial velocity of 5 m/s. ten second later, he is moving at 15 m/s. what is his

pantera1 [17]

$\Large {{ \sf {Question :}}}$

<h3>A boy who is riding his bicycle, moves with an initial velocity of 5 m/s. Ten second later, he is moving at 15 m/s. What is his acceleration?</h3>

$\Large {{ \sf {Given :}}}$

<h3>Initial Velocity (<em>u</em>) - 5 m/s</h3><h3>Final Velocity (<em>v</em>) - 15 m/s</h3><h3>Time (<em>t</em>) - 10 sec</h3>

$\Large {{ \sf {Formulae :}}}$

<h3>If the velocity of an object changes from an initial value <em>u </em>to the final value <em>v </em>in time <em>t,</em><em> </em>the acceleration <em>a</em> is, </h3><h3> $a \: = \frac{v - u}{t}$ </h3><h3> $\Large {{ \sf {Step-by-step explanation :}}}$ </h3>

$a \: = \frac{v - u}{t} \\ or \: \: a = \frac{(15 - 5)}{10} m \: s^{ - 2} \\ or \: \: a \: = \frac{10}{10}m \: s^{ - 2} \\ or \: \: a = 1m \: s^{ - 2}$

$\Large {{ \sf {Answer :}}}$

<h3>His acceleration is </h3><h3> $1m \: s^{ - 2}$ </h3><h3 /><h3 />

5 0

2 years ago

Any four difference between velocity and acceleration​

Any four difference between velocity and acceleration