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trasher [3.6K]
4 years ago
5

A runner accelerates to a velocity of 4.15m/s due west in 1.50 s. his average accelerates is 640m/

Physics
1 answer:
Ivanshal [37]4 years ago
7 0
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A model rocket rises with constant acceleration to a height of 4.2 m, at which point its speed is 27.0 m/s. How much time does i
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Answers:

a) t=0.311 s

b) a=86.847 m/s^{2}

c) y=1.736 m

d) V=17.369 m/s

Explanation:

For this situation we will use the following equations:

y=y_{o}+V_{o}t+\frac{1}{2}at^{2} (1)  

V=V_{o} + at (2)  

Where:  

y is the <u>height of the model rocket at a given time</u>

y_{o}=0 is the i<u>nitial height </u>of the model rocket

V_{o}=0 is the<u> initial velocity</u> of the model rocket since it started from rest

V is the <u>velocity of the rocket at a given height and time</u>

t is the <u>time</u> it takes to the model rocket to reach a certain height

a is the <u>constant acceleration</u> due gravity and the rocket's thrust

<h2>a) Time it takes for the rocket to reach the height=4.2 m</h2>

The average velocity of a body moving at a constant acceleration is:

V=\frac{V_{1}+V_{2}}{2} (3)

For this rocket is:

V=\frac{27 m/s}{2}=13.5 m/s (4)

Time is determined by:

t=\frac{y}{V} (5)

t=\frac{4.2 m}{13.5 m/s} (6)

Hence:

t=0.311 s (7)

<h2>b) Magnitude of the rocket's acceleration</h2>

Using equation (1), with initial height and velocity equal to zero:

y=\frac{1}{2}at^{2} (8)  

We will use y=4.2 m :

4.2 m=\frac{1}{2}a(0.311)^{2} (9)  

Finding a:

a=86.847 m/s^{2} (10)  

<h2>c) Height of the rocket 0.20 s after launch</h2>

Using again y=\frac{1}{2}at^{2} but for t=0.2 s:

y=\frac{1}{2}(86.847 m/s^{2})(0.2 s)^{2} (11)

y=1.736 m (12)

<h2>d) Speed of the rocket 0.20 s after launch</h2>

We will use equation (2) remembering the rocket startted from rest:

V= at (13)  

V= (86.847 m/s^{2})(0.2 s) (14)  

Finally:

V=17.369 m/s (15)  

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