Net force to stop a truck and car acceleration

[thua]

1. A 1.2 x 10^4 kg truck is traveling south at 22 m/s.
a) What net force is required to bring the truck to a stop in 330 m?
b) What is the cause of this net force?
Answer: a) 8.0 x 10^3 N [N]

2. Each of the four wheels of a car pushes on the road with a force of 4.0 x 10^3 N [down]. The driving force on the car is 8.0 x 10^3 N [W]. The frictional resistance on the car is 6.0 x 10^3 N [E]. Calculate the following:
a) the mass of the car
b) the car's acceleration
Answer: a) 1.6 x 10^3 kg b) 1.2 m/s^2 [W]

I have the answers to these two questions, but I really need help on how to understand and do it.

So far...
1. a) v = d/t
t = d/v
t = (0.33 km)/(22 m/s )
t = 0.015 s

a = v/t
a = (22 m/s )/(0.015 s)
a = 1466.7 m/s^2 ?? <====== stopped here because this makes no sense..

 

[Seiya]

the time makes no sense... you are right :p

here is your problem

t = (0.33 km)/(22 m/s )

the distance has to be in meters...

 

[whozum]

1. A 1.2 x 10^4 kg truck is traveling south at 22 m/s.
a) What net force is required to bring the truck to a stop in 330 m?
b) What is the cause of this net force?
Answer: a) 8.0 x 10^3 N [N]

If you've covered energy then that would be the best way to solve the problem. If not then the following equation should look familiar:

$v_f^2 = v_i^2 + 2ad$ which is what you should use to tackle this problem.

 

[thua]

whozum... is that to find the acceleration?

(22 m/s )^2 = (0 m/s )^2 + 2a(330 m)
a = 0.73 m/s^2

f = ma
f = 8760 N <----- something's still wrong

 

[Seiya]

if you want the truck to stop acceleration has to be negative...

 

[thua]

that still doesn't work though because the net force comes up to -8670 N .

i think I'm totally missing out on something here...

 

[whozum]

that still doesn't work though because the net force comes up to -8670 N .

The acceleration is not south but north, -8670 N = 8670N [N]. However I don't know why they have 8.0 x 10^3N as the answer, your answer of 8670 is alomst correct.

By the way the correct solution is

$v_f^2 = v_i^2 + 2ad$ you had the velocities mixed up.

$(0)^2 = (22)^2 + 2a(330m)$

[itex]a = \frac{22^2}{-660} = -0.733333 [/tex]<br /> <br /> $F = ma = m(-0.7333333) = 8800N = 8.8 \times 10^3 N [N]$[/itex]

 

[thua]

okay thankssss... and for number 2... i really don't know how itll work cause mass is equal to net force over acceleration, but all i have is net force... no acceleration... so how would that work out?

 

[whozum]

The force from the 4 tires is the weight of the car (normal force). You can find the mass from there.

 

[thua]

does that mean that the answer is wrong again? because i came up with 16000 N which is 1.6 x 10^4 and they say its 1.6 x 10^3 kg.

and does 1 N equal to 1 kg... I am not very familiar with that.
but 1 N actually equals to 1 kg x m/s^2, though right?

 

[Seiya]

you answered your own question a Newton is a kg * m/s^2