Calculating Velocity and Position with Constant Acceleration

In summary, 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. Using the definition of acceleration, a=dv/dt, the velocity at t=3.0 s is calculated to be 10.6 m/s. In addition, the position of the car is found by integrating the velocity and using the given condition x=5.0 m at t=1.0 s, resulting in a position of approximately 20 m at t=3.0 s.
  • #1
Warcus
4
0
Hi there everyone! I'm a first-year physics student and I need some help with a couple of questions! OK, here it goes :)

Homework Statement



Task 1​
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)

There are four possible answers:
(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

Task 2​
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

There are four possible answers:
(a) x (3) = 15m
(b) x (3) = 18m
(c) x (3) = 20m
(d) x (3) = 23m

Homework 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)

The Attempt at a Solution



Task 1​
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

Task 2​

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
 
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  • #2
Warcus said:

Homework Statement



Task 1​
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)
Hi Warcus. Welcome to PF. :smile:

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 :-p.

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
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.. :confused:

Could you try working it out step-by-step so I can see how it is done? :rolleyes:
 
  • #4
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
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 = 4,3V(t) = (1.4*t2)/2 + C
V(3) = (1.4*32)/2 + 4,3
V(3) = 10,6
V(3) ≈ 11
V(3) = answer number d

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! :biggrin:
 
  • #6
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.

[tex]x=\int{v(t)dt}=\int{(0.7t^2+4.3)dt}[/tex]

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
Alright, task 2:

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
x ≈20 → Answer C

Thanks for helping me get the answers and teaching me to antideriviative! :biggrin: :biggrin:
 

Related to Calculating Velocity and Position with Constant Acceleration

1. What is acceleration?

Acceleration is a measure of how quickly an object's velocity changes over time. It can be calculated by dividing the change in velocity by the change in time.

2. How is acceleration related to distance?

Acceleration and distance are related through the equation d = 1/2at^2, where d is the distance traveled, a is the acceleration, and t is the time.

3. How does acceleration affect an object's motion?

Acceleration can either increase or decrease an object's speed, or change its direction of motion, depending on the direction and magnitude of the acceleration.

4. What is the difference between acceleration and deceleration?

Acceleration refers to an increase in velocity, while deceleration refers to a decrease in velocity. However, both involve a change in an object's speed over time.

5. How can acceleration be measured?

Acceleration can be measured using various tools such as accelerometers, speedometers, and motion sensors. It can also be calculated using mathematical equations.

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