# Acceleration and distance!

1. Aug 31, 2012

### Warcus

Hi there everyone! I'm a first-year physics student and I need some help with a couple of questions! OK, here it goes :)

1. The problem statement, all variables and given/known data

The acceleration of a vehicle is given by a(t) = αT, where α=1.4 m/s^3. At the time t=1.0 s
the car has a speed of 5.0 m/s, what is the velocity at t=3.0 s?

(α = alpha) (In other words a(t)=α=1.4 m/s^3 is not a typo)

(a) y (3) = 6.3 m / s
(b) v (3) = 7.4 m / s
(c) y (3) = 8.7 m / s
(d) y (3) = 11 m / s

We use with the same car as in the previous problem. We learn that the car
position at t = 1,0s is 5.0 m Find the car's position at t = 3.0 s

(a) x (3) = 15m
(b) x (3) = 18m
(c) x (3) = 20m
(d) x (3) = 23m

2. Relevant equations

The relevant equations would be the kinematic equations for constant acceleration:
V = Vo + at
X = Xo + VoT+ 0.5aT^2
V^2=Vo^2 + 2a(-Xo)

3. The attempt at a solution

V = Vo + at

V = 5 m/s + 1,4 m/s^2 *2 s (the difference between the two t-values is 2)
V = 7,8 m/s

This answer is not one of the possible solutions --> I have done something wrong
Possible solutions (again):
(a) y (3) = 6.3 m / s
(b) v (3) = 7.4 m / s
(c) y (3) = 8.7 m / s
(d) y (3) = 11 m / s

X=VoT + 0,5at^2

X=5 m/s x 2 s + 0,5*1,4 m/s^2 * (2s)^2 (Again, the difference between the t-values = 2)
X = 12,8 m

The task specifies that the distance at T(1)=5 --> Total distance = 5 m + 12,8 m = 17,8

Again, ther are four possible answers, and I'm fairly sure that the answer is b:
(a) x (3) = 15m
(b) x (3) = 18m
(c) x (3) = 20m
(d) x (3) = 23m
Nevertheless, I am a bit unsure about my calculation, and posted this Task as well :)

Thanks a lot for your help! :D

2. Aug 31, 2012

### ehild

Hi Warcus. Welcome to PF.

In the formula for the acceleration you meant t instead of T I think.

So a(t)=αt, and α=1.4 m/s3, so a(t)=1.4t.
This is the case of non-uniform acceleration, your relevant equations are irrelevant :tongue2:.

Go back to the definition of acceleration: it is the time derivative of velocity. a=dv/dt. The opposite is also true: the velocity is obtained by integrating the acceleration: v(t)= ∫a(t)dt, and adding an arbitrary constant. The constant can be found from the condition that v=5.0 m/s at t=1.0 s.

ehild

3. Sep 1, 2012

### Warcus

Hi and thanks for the assistance ehild! I tried doing what you said, but as this is my first year studying physics I encountered some problems... Take a look, will you?

v(t) = ∫a(t)dt
v(t) = (t^a) * t + C (where a is 1,4t)
v(3) = (3^1,4*3) * 3 + C = 9,12 + C

Vo = V(1) = 1,4(1) + C --> C = 3,6

v(3) = 12,72

which can't be right according to the answers the teacher gave me..

Could you try working it out step-by-step so I can see how it is done?

4. Sep 1, 2012

### ehild

The acceleration is a=1.4 t. v(t) is the integral or primitive function of the acceleration. V(t)=∫(1.4 t dt) = 1.4 ∫(t dt) =1.4 t2/2+C. You have studied some calculus , about integration, have you not?
Now substitute t=1 and v=5 and you get C.
Use that value of C to get the velocity at t=3 s.

ehild

5. Sep 1, 2012

### Warcus

V(t) = (1.4*t2)/2 + C
C = V(t) - (1.4*t2)/2
C = V(1) - (1.4*12)/2
C = 5 - (1,4 *12)/2
C = 4,3

V(t) = (1.4*t2)/2 + C
V(3) = (1.4*32)/2 + 4,3
V(3) = 10,6
V(3) ≈ 11

And yeah, I studied some integration, but hey... It was three years ago and was probably just the basics. Anyways, thanks for the help, it was greatly appreciated!

6. Sep 1, 2012

### ehild

Do not stop yet. Task 2 says that it is the same car as before, which has the velocity (you know it already) V(t) = (0.7*t2 + 4.3. You need the position as function of time. It is the integral (antiderivative) of the velocity.

$$x=\int{v(t)dt}=\int{(0.7t^2+4.3)dt}$$

Refresh your knowledge about the integrals (antiderivatives) If it is too difficult, you can cheat a bit.

http://calculator.tutorvista.com/math/585/antiderivative-calculator.html#
When you know the expression for x(t), you can find the unknown C again from the condition x=5.0 m at t=1.0 s.

ehild

7. Sep 1, 2012

### Warcus

x = ∫v(t)dt = ∫(0.7t2+4.3)dt = (0.7t3)/3 + 4.3t + C

x = (0.7t3)/3 + 4.3t
x(1) = (0.7*13)/3 + 4.3*1 + C
5 = (0.7*13)/3 + 4.3*1 + C
C = 5 - 4.53333... = 0.4666...

x = (0.7*33)/3 + 4.3*3 + C
x = (0.7*33)/3 + 4.3*3 + 0.46666...
x = 19.6333