Question

A 873 kg dragster, starting from rest, attains a speed of 26.3 m/s in 0.59 s.
a) Find avg. acceleration: 45 m/s/s. FOUND
b) Size of avg. force: 39000 N. FOUND
c) Assume the driver has a mass of 68 kg. What horizontal force does the seat exert on the driver? 3100 N. FOUND
d) Is the driver's mass in part c) an inertial or gravitational mass?
---
A 65 kg swimmer jumps off a 10 m tower.
a) find the swimmer's velocity when hitting the water.
b) The swimmer comes to a stop 2 m below the surface. Find the net force exerted by the water.

Answer 1

ok start solving.. what u waiting for? :P

Answer 2

Wait! editing

Answer 3

Have you seen kinetic and potential energy?

Answer 4

@Lessis: No, I haven't ... we just started this topic ...

Answer 5

I need letter c) for the first one, and I have 3 more problems until the second one.
Kinetic is energy in motion, right?

Answer 6

\[E_{K} = \frac{ 1 }{ 2 }mv^{2} \]

Answer 7

... I don't recognize that formula...

Answer 8

That's Kinetic Energy.

Answer 9

Well, yes I know that, but we haven't learned it.
So we shouldn't be using it, right?

Answer 10

You can use it, but I'm pretty sure it's not recommended.
Do you know your vertical throw equations?

Answer 11

Eh. What...? o.o

Answer 12

All I know are these:\[F = ma\] \[\Sigma Fy = Fa - W = ma\]

Answer 13

And the two about acceleration and velocity that you told me.

Answer 14

You can calculate the velocity of the swimmer using the equation I gave you for constant acceleration and initial velocity cero.

Answer 15

cero?

Answer 16

After you calculate the velocity with which the swimmer touches the water, you can calculate his deceleration, considering that he stopped after two meters. (You need to calculate the acceleration required to get to a complete stop, from the velocity you calculated, to a velocity of 0, in just two meters)
After you get that acceleration, you can calculate the force by multiplying by mass.

Answer 17

Is that Spanish for zero?

Answer 18

I haven't gotten to that question yet

Answer 19

... Perdón. xD Fue una ligera falla de ortografía dado a que estoy acostumbrado (malacostumbrado?) a escribir cero. :P

Answer 20

& I think I'd need a diagram...
no me gusta espanol = n=

Answer 21

wait I have another question! o.o (see pulse)

Answer 22

.... nvm i put it here
After a day of testing race cars, you decide to take your own 1550 kg car onto the test track. While moving down the track at 10 m/s you suddenly accelerate to 30 m/s in 10 s. What is the avg. net force you have applied to the car during the 10s interval?

Answer 23

... lol "dun dun dunn" ??
good pic I get it

Answer 24

\[Average Acceleration = \frac{ \Delta V }{ t }=\frac{ V_{2} - V_{1} }{ t } = \frac{ 30 \frac{ m }{ s } - 10\frac{m}{ s } }{ 10 s }\]
\[Average Force = mass*averageacceleration\]

Answer 25

wut

Answer 26

\[a_{avg} = \frac{ \Delta v }{ t } = \frac{ v _{2} - v _{1}}{ t } = \frac{ 30 m/s ^{2} - 10 m/s ^{2}}{ 10s }\]
\[F _{avg} = m \times a _{avg}\]

Answer 27

that's my idea of it

Answer 28

Yes. But the velocities are given in meter per second, not meter per second squared.

Answer 29

-.- Grr.
I forgot it was velocity & not acceleration.

Answer 30

Does that equal 20 m then?

Answer 31

2 m/s/s*

Answer 32

The difference in speed is of 20 meters per second, and the acceleration is of 2 meters per second squared, yes.

Answer 33

#thumbsup

Answer 34

A race car has a mass of 710 kg. It starts from rest and travels 40 m in 3.0 s. The car is uniformly accelerated during the entire time. What net force is applied to it?
Answer is 6300 N.

Answer 35

I get the velocity\[v \approx 13.3 m/s\]

Answer 36

This becomes \[a _{avg} = \frac{ v }{ t } = \frac{ 13.3 }{ 3.0 } \approx 4.44 m/s ^{2}\]Now I plug this in\[F = ma = 710 \times 4.44 \approx 3100 N\]
I got the answer, I think.

Answer 37

Except a is not average...
Did I do it right?
@Lessis

Answer 38

Wait.... wtf NO I didn't get the answer = n =

Answer 39

You need an equation I don't remember how to get. D:
\[x =v_{0}t + \frac{ 1 }{ 2 }at^{2}\]

Answer 40

it's 4.43 not 4.44

Answer 41

... darn I hate those equations!
Okay:
\[x = v _{0}t + \frac{ 1 }{ 2 }at\]becomes\[40 = 0 + \frac{ 1 }{ 2 }a \left( 3.0 \right)\]which translates into\[80 = 3.0a\]and that makes\[a \approx 26.67\]which could be 27 or 26.7 depending on sig figs

Answer 42

Time is squared.

Answer 43

m/s/s since I forgot.
... Crap.
\[80 = 9.0a\]\[a \approx 8.89\] depending on sig figs, could be 8.9 or 9

Answer 44

\[8.9 m/s ^{2}\] it is.
\[F = ma = 710 \times 8.9 \approx 6300 N\]

Answer 45

Correct!

Answer 46

Thanks! :)
A force of -9000 N is used to stop a 1500 kg car at 20 m/s. What braking distance is needed to bring the car to a halt?

Answer 47

For the swimming problem, you need the same equation to calculate the force exerted by the water, by the way.

Answer 48

Ok, I gotcha. :p
Do I still use F= ma?

Answer 49

For the last one I posted*

Answer 50

\[F =ma\]\[-9000 = 1500a\]\[a = -6\]

Answer 51

Yes.

Answer 52

I'm guessing I find time and then use the equation to find \[\Delta x\]

Answer 53

Use
\[v = v_{0} + at\]
To calculate the time needed for a complete stop.
Then use
\[x = v_{0}t + \frac{ 1 }{ 2 }at^{2} \]
To calculate the distance.

Answer 54

Okay.
\[v = v _{0} + \frac{ 1 }{ 2 }at ^{2}\]

Answer 55

That makes\[20 = 0 + \frac{ 1 }{ 2 }\left( -6 \right)t ^{2}\]
am I right

Answer 56

Wrong equation. You typed v instead of x.
And V0 is your starting speed, which in this case is 20 m/s.

Answer 57

Wait.... I did the wrong equation completely #lolwhat

Answer 58

\[v = v_{0} + at\]
\[0 = 20 \frac{ m }{ s }+ at\]

Answer 59

\[20 = 0 + -6t\]

Answer 60

.-. oh right. \[0 = 20 m/s - 6t\]which is\[t = \frac{ -20 }{ -6 } \approx 3.3 s\]

Answer 61

\[x = v_{0}t + \frac{ 1 }{ 2 }at^{2}\]
x = (20*3.3) + 0.5*(-6)*(3.3)*(3.3)

Answer 62

\[x \approx 71 m\]wrong,it's 33 -.-

Answer 63

i messed up on the calculations

Answer 64

somewhere.... even though i use a calculator...

Answer 65

https://www.google.com.mx/search?q=%2820*3.3%29+%2B+0.5*%28-6%29*%283.3%29*%283.3%29&ie=utf-8&oe=utf-8&rls=org.mozilla:en-US:official&client=firefox-a&gws_rd=cr&ei=DqRzUs6cNu7ksATEk4DwBA

Answer 66

oh i left out the -6

Answer 67

ok time for swimmer problem

Answer 68

i got the pic

Answer 69

.-. i need time to solve///

Answer 70

I mean, I need t to solve

Answer 71

\[x = v_{0}t + \frac{ 1 }{ 2 }at^{2}\]

Answer 72

well yes

Answer 73

In this case, x is -10 meters. (You've fallen ten meters from the top of the pool)
V0 is 0. (The speed at which you start, when you're at the top)
a is g, which is equal to 9.8 m/s.

Answer 74

a is also negative, considering how you're falling, by the way.

Answer 75

ok
i got approx. 1.43

Answer 76

That is your time, yes.
Now, considering how the acceleration is constant use:
\[a = \frac{ v-v_{0} }{ t } = \frac{ v }{ t }\]
To calculate the speed.

Answer 77

Remember that a is equal to g that is equal to -9.8 meters per second squared.

Answer 78

ok

Answer 79

9.8 = v/1.43?

Answer 80

I get 14.014

Answer 81

Yes. Just remember that a was negative, so v must be negative. The negative velocity means that you're falling.

Answer 82

Vf=14 then?

Answer 83

Yes.

Answer 84

Now, you know that after touching the pool, he falls for two meters and then goes to a complete stop. So you have the next equation for the average acceleration in the water.

\[a = \frac{ \Delta V }{ t }= \frac{ V_{2} - V_{1} }{ t } = \frac{ -V_{1} }{ t }\]
Where v1 is the speed you just calculated, -14 m/s.

So.

\[a = -\frac{ -14 m/s }{ t } = \frac{ 14 m/s }{ t }\]

Answer 85

You also know that:
\[x = v_{0}t + \frac{ 1 }{ 2 }at^{2}\]
Where V0 is your -14 m/s and x is -2 m.

Answer 86

\[-2m = (-14m/s) t + \frac{ 1 }{ 2 }at^{2}\]

Answer 87

So, now you have two equations and two variables (a and t). So you have to solve the system.

Answer 88

o.o I come back to see this..

Answer 89

wait

Answer 90

ok I got it, thanks @Lessis

Answer 91

#offline