Find distance during acceleration

[Billy G]

a rocket moves from 30m/s to 120m/s accelerating at a constant rate of 18.0 m/s^2. How far does the rocket travel during acceleration.
Here is what I figured.
init Vel (iV)= 30m/s
final vel (V)= 120m/s
acel (a)= 18 m/s^2
d=?
this is what i figured and I am confused perhaps over thinking this.
formula- V^2=iV^2+2(a)(d)
120^2=30^2 + 2(18) (d)
14400=900 + 36(d)....... now not sure if it should be d^2 or 2x d (0) is zero or just (d)
14400=936(d)
936 d= 15.38m and if it should be squared then it is 236.5m.

Complete brain fart, and totally confused on this one. I am thinking it is 15.38m rounded up to 15.4m. but uncertain...
Please advise. Thanks.

 

[Romsek]

if all else fails you can always just integrate twice

\(\displaystyle a(t) = 18,~t\in [0, 5]\\
v(t) = \displaystyle \int ~a(t) +v_0 = 18t+30 \\
d = \displaystyle \int_0^5 v(t) ~dt = \\

\left . 9t^2 +30t \right |_0^5 = 375 m\\\)

Hrm. Maybe you don't know about integration.

Since the acceleration is constant you can find the average speed of the rocket and use that.

\(\displaystyle v_{avg} = \dfrac{30+120}{2} = 75 ~m/s\)

Now the rocket travels the distance due to this average speed over 5 seconds

\(\displaystyle d = 75~m/s \cdot 5~s = 375 m\)

 

[Billy G]

Romsek
not sure I am following you on above. where did you come up with t=5. i tried working thru the problem again so i can see it and get it..
i got same for average velocity/speed.. but i used for time calculation the following
final velocity (120)=init velocity (30) + Acceleration (18) Time (X).
120=48t= 2.5s. then I inserted into the d=avgVelocity (75) X time (2.5)= d= 187.5m.
48

what am i missing here. just using the variables given. The integration thing is way over my head..just using numbers given... the suggestion of the average speed makes sense.
Thanks.

 

[Deleted member 4993]

a rocket moves from 30m/s to 120m/s accelerating at a constant rate of 18.0 m/s^2. How far does the rocket travel during acceleration.
Here is what I figured.
init Vel (iV)= 30m/s
final vel (V)= 120m/s
acel (a)= 18 m/s^2
d=?
this is what i figured and I am confused perhaps over thinking this.
formula- V^2=iV^2+2(a)(d)
120^2=30^2 + 2(18) (d)
14400=900 + 36(d)....... now not sure if it should be d^2 or 2x d (0) is zero or just (d)
14400=936(d)
936 d= 15.38m and if it should be squared then it is 236.5m.

Complete brain fart, and totally confused on this one. I am thinking it is 15.38m rounded up to 15.4m. but uncertain...
Please advise. Thanks.

Here is what I figured.
init Vel (iV)= 30m/s
final vel (V)= 120m/s
acel (a)= 18 m/s^2
d=?
this is what i figured and I am confused perhaps over thinking this.
formula- V^2=iV^2+2(a)(d)
120^2=30^2 + 2(18) (d)
14400=900 + 36(d) ....................................... correct upto here

14400 - 900 = 36 * d ..................................................isolating 'd'

13500 / 36 = d

d = 375 m

However, your accompanying questions worry me. Why would you think:

now not sure if it should be d^2

or

2x d (0)

You may want to sit down face-2-face with your instructor and ask these questions. If you want, we could try to answer those here (over the computer screen) but that might be a bit difficult! But we are ready.....

 

[Deleted member 4993]

Romsek
not sure I am following you on above. where did you come up with t=5. i tried working thru the problem again so i can see it and get it..
i got same for average velocity/speed.. but i used for time calculation the following
final velocity (120)=init velocity (30) + Acceleration (18) Time (X).
120=48t= 2.5s. then I inserted into the d=avgVelocity (75) X time (2.5)= d= 187.5m.
48
How did you get 48t?
what am i missing here. just using the variables given. The integration thing is way over my head..just using numbers given... the suggestion of the average speed makes sense.
Thanks.

final velocity (120)=init velocity (30) + Acceleration (18) Time (X).
120=48t= 2.5s. then I inserted into the d=avgVelocity (75) X time (2.5)= d= 187.5m.
48
How did you get 48t? Did you add 30 to 18t? That would be like adding apples to roses - not allowed!!!

 

[Steven G]

a rocket moves from 30m/s to 120m/s accelerating at a constant rate of 18.0 m/s^2. How far does the rocket travel during acceleration.
Here is what I figured.
init Vel (iV)= 30m/s
final vel (V)= 120m/s
acel (a)= 18 m/s^2
d=?
this is what i figured and I am confused perhaps over thinking this.
formula- V^2=iV^2+2(a)(d)
120^2=30^2 + 2(18) (d)
14400=900 + 36(d)....... now not sure if it should be d^2 or 2x d (0) is zero or just (d)
14400=936(d)
936 d= 15.38m and if it should be squared then it is 236.5m.

Complete brain fart, and totally confused on this one. I am thinking it is 15.38m rounded up to 15.4m. but uncertain...
Please advise. Thanks.

Since when is 900 + 36d = 936d.
Consider this problem. 600 + 1d where d= 1million. So you are saying that 600 + 1million = 601 million. That is if you have $600 and some loans you $1 million, now you have $601 million. So after paying back the $1 million you now have $600 million. All this money from just $600.