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[LuizH]

There is information missing from the question.

What about if he started accelerating slowly then all of a sudden went 100% throttle during the last 5 seconds to reach 30m/s? The distance traveled would be drastically different than if he accelerated at a constant rate. I would think the question is assuming constant rate of acceleration and deceleration, because its much easier to figure out. Otherwise, the distance travelled could be much different.

Well I assume constant acel and decel, most of the times the problem would say not to asume.

[LuizH]

Another way to find the lenght would be by calculating the area under the graph VxT

[magnusson]

Another way to find the lenght would be by calculating the area under the graph VxT

ok that is how i am suppose to do it, all the stuff you said before, dint understand 1 bit of it(i'm stupid i know).

first part of the question is to draw the trip of the car.

[LuizH]

So the first part of trip is the constant aceleration
V goes from 0 to 30 in 20s, so the you got 2 points (0,0) and (20,30) since aceleration is constant you join the two points by a straight.
Now second part of trip accelaration=0 so V is constant and its a straight till t=50s
The last part V goes from 30 to 0 in 30s so you got the point (50,30) and (80,0) and since decelaration(brake) is constant you joint the points by a straight but this time going down.

[LuizH]

double post, this one is the correct graph with t on the x axis

[ShackAttack12]

I was just posting to you check my calcs because i tend to make silly mistakes, specialy on vacations when i realized the lenght of first part is 300 not 100
So total lenght is 1650
that changes average speed too Vav=20,625

Let me work this out to see if i come up with the same thing.

First, list the givens:
Acceleration:
Initial Velocity = 0m/s
Final Velocity = 30m/s
Time= 20 seconds
Distance, x = ???

Assuming constant acceleration rate:

Average Acceleration: ( V - Vo ) / time
Average Acceleration: ( 30m/s - 0m/s ) / 20s
Average Acceleration: 1.5m/s^2

To find distance:
x = Vot + 1/2at^2
x = (0*20) + (1/2)(1.5m/s^2)(20^2)
x = 300m

For travelling at a constant rate, that's easy.
He went 30m/s for 30 seconds.
Distance = Velocity * Time
x = 30m/s * 20s
x = 600m

For the Deceleration, you just do the opposite of acceleration, since its essentially negative acceleration:
Vo = 30m/s
V = 0m/s
t = 30s
x = ???

Average Decel, a = - ( V - Vo ) / time
Average Decel, a = - ( 0m/s - 30m/s ) / 30s
Average Decel, a = 1m/s^2

To find Distance:
x = Vt + 1/2at^2
x = (0m/s)(30s) + (1/2)(1m/s^2)(30s^2)
x = 450m

So add the 3 values:
300m + 600m + 450m = 1350m

2, Average velocity is distance over time.
x = v * t
1350m = v * (20s+30s+30s)
1350m / 80s = v
v= 16.875m/s

[LuizH]

you made
x = 30m/s * 20s
x = 600m
where should be 30 s
the rest is just like I've made other then more organized

[ShackAttack12]

science class question's. giggity

car setting off from a standstill point. it takes him 20 seconds to catch the speed 30 meters per seconds (30x3.6=108km/h right?). Then he drives on a even speed for the next 30 seconds. and then he brakes and it takes him 30 seconds to stop.

Q1; What is the cars trip in length?

Q2; what is his average speed?

need to know how the fuck i answer there damn question, i knew how but i decided not to remember...

[ShackAttack12]

you made
x = 30m/s * 20s
x = 600m
where should be 30 s
the rest is just like I've made other then more organized

Oh shit, you are right. LOL

[LuizH]

science class question's. giggity

Then he drives on a even speed for the next 30s

[ShackAttack12]

Here's the answer:

First, list the givens:
Acceleration:
Initial Velocity = 0m/s
Final Velocity = 30m/s
Time= 20 seconds
Distance, x = ???

Assuming constant acceleration rate:

Average Acceleration: ( V - Vo ) / time
Average Acceleration: ( 30m/s - 0m/s ) / 20s
Average Acceleration: 1.5m/s^2

To find distance:
x = Vot + 1/2at^2
x = (0*20) + (1/2)(1.5m/s^2)(20^2)
x = 300m

For travelling at a constant rate, that's easy.
He went 30m/s for 30 seconds.
Distance = Velocity * Time
x = 30m/s * 30s
x = 900m

For the Deceleration, you just do the opposite of acceleration, since its essentially negative acceleration:
Vo = 30m/s
V = 0m/s
t = 30s
x = ???

Average Decel, a = - ( V - Vo ) / time
Average Decel, a = - ( 0m/s - 30m/s ) / 30s
Average Decel, a = 1m/s^2

To find Distance:
x = Vt + 1/2at^2
x = (0m/s)(30s) + (1/2)(1m/s^2)(30s^2)
x = 450m

So add the 3 values:
300m + 900m + 450m = 1650m

2, Average velocity is distance over time.
x = v * t
1650m = v * (20s+30s+30s)
1650m / 80s = v
v= 20.625m/s

[magnusson]

i'm starting to understand this much better thanks for your time. and for those ho have nothing to do here's another question


ROCKET!! its weight is 6000kg and its shot upp whit the force 90000N

Q1 does she fly?

Q2 what will her acceleration be?

[Shadow]

F-G=ma where F = 90 000N, G=mg where m = 6000kg and g = 9,81 m/s^2

=> a= (F-G)/m = (90 000N - 6000kg*9,81m/s^2)/(6000kg) = 5,19 m/s^2

So, yes, she'll fly.

[Shadow]

Provided that the thrust force is a constant and it's not let off. If it's an impulse, then it won't fly very far.

[LuizH]

Dang I jut whent to brush my theets before answer that and shadow allready did