Kinematics Maths Problem

[HUNTER.P]

Been stumped in this for ages now. Ain't got a clue how to go about it

 

[doogs]

The answer is 5.93 (I think).

If S = distance, T = time, you are given T=1.8
Therefore, S = (2 x 1.8 x1.8) + (4 x 1.8) -3 = 10.68
speed = distance / time
speed = 10.68 / 1.8 = 5.933333

 

[mufc802]

s=2t^2+4t-3 ... t=1.8

S = Distance
T = Time
V = Velocity

s=2(1.8)^2+4(1.8)-3 ... s=10.68m

s=vt Distance=Velocity*Time (this is a common formula you should know)

v=s/t
10.68/1.8 = 5.93ms^-2

 

[HUNTER.P]

Jeez i was over thinking that too much. Cheers guys

 

[Amleto]

erm, dont you differentiate?

Question is what is velocity AT time t=whatever. NOT what is the average velocity after time t=x, which is what the other guys have done.

v(t) = 4t + 4

e: mufc, velocity is m/s not m/s/s

 

[Participant]

Is it a ten mark question? Got to be more to it than speed = distance / time.

 

[Amleto]

Is it a ten mark question? Got to be more to it than speed = distance / time.

there is.

 

[Smoking Man]

OcUK experts strike again.

 

[HUNTER.P]

True, does me think there is more involved for 10 marks now.

erm, dont you differentiate?

Question is what is velocity AT time t=whatever. NOT what is the average velocity after time t=x, which is what the other guys have done.

v(t) = 4t + 4

e: mufc, velocity is m/s not m/s/s

Any chance you could show me the method? Got another few questions like this.

 

[paulc25]

Differentiating polynomials couldn't be easier, see here:

http://www.wikihow.com/Differentiate-Polynomials

 

[m3csl2004]

velocity = rate of change of position, differentiate and solve

4t+4 when t=1.8

11.2?

 

[Liampope]

velocity = rate of change of position, differentiate and solve

4t+4 when t=1.8

11.2?

Was just gonna post exactly that myself

 

[Judgeneo]

velocity = rate of change of position, differentiate and solve

4t+4 when t=1.8

11.2?

Bingo.

 

[danza]

11.2m/s is correct.

 

[Dunks]

s=2t^2+4t-3

differentiate respect to t, ds/dt=4t-4

if t=1.8

ds/dt=11.2

 

[Psiko]

s=2t^2+4t-3 ... t=1.8

S = Distance
T = Time
V = Velocity

s=2(1.8)^2+4(1.8)-3 ... s=10.68m

s=vt Distance=Velocity*Time (this is a common formula you should know)

v=s/t
10.68/1.8 = 5.93ms^-2

This assumes a constant velocity which is not correct in this question. As others have stated, you need to differentiate to find the velocity.

 

[TheOverLord]

I think using the letter 's' to describe postion and not speed threw a few people off.

Examiners just looove to be sneaky like that

 

[mufc802]

This assumes a constant velocity which is not correct in this question. As others have stated, you need to differentiate to find the velocity.

I see, I've never come across needing to use differentiation in any of the M1 stuff I did.

 

[DaveF]

I think using the letter 's' to describe postion and not speed threw a few people off.

Examiners just looove to be sneaky like that

's' is actually the "standard" letter for distance (where you're not wanting to imply distance in the horizontal direction).

("standard" in quotes because it's not set in stone. That is, it's a lot less standard than t for time or v for velocity).

 

[Alex74]

's' is actually the "standard" letter for distance (where you're not wanting to imply distance in the horizontal direction).

("standard" in quotes because it's not set in stone. That is, it's a lot less standard than t for time or v for velocity).

This. Especially when you start getting to more interesting physics where you define line elements by ds, it's the generally accepted notation