How Does Constant Acceleration Affect Distance Traveled?

In summary: So, a is the acceleration, which is at 3 metres per second-squared, the initial velocity is 0, because the question says it starts "from rest", Δt, this will determine the interval of time used in determining the velocity. v2 and Δd is unknown.Then, using the equation for displacement, you can find that the displacement is 0, and that the object has traveled a total of 3 metres.
  • #1
jbeannie05
10
0
A car accelerates at 3 meters squared from rest for 10s. How far does it travel?

a = 3 meters squared
v1 = 0?
v2 = ?
Δd = ?
Δt = 10s

Why is velocity 1, 0? Why is delta t 10 seconds?
 
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  • #2
welcome to pf!

hi jbeannie05! welcome to pf! :wink:
jbeannie05 said:
Why is velocity 1, 0? Why is delta t 10 seconds?

v1 is the initial velocity (the starting velocity), and since the question says it starts "from rest", that means v1 = 0

∆t is the time from start to finish: since it says it accelerates "for 10s", that means ∆t = 10

now apply the usual constant acceleration equations …

what do you get? :smile:

(btw, it's not 3 metres squared, it 3 metres per second-squared, 3 ms-2)
 
  • #3
v1 is given to you as 0 because the car starts from rest
Δt is also given to you: the car accelerates during 10s
With this you must have seen the formulas to deduce v2 (the velocity after 10s) and the total distance travelled.
 
  • #4
Δd=v1Δt+aΔt^2

This should answer your question.
 
  • #5
gbaby370 said:
Δd=v1Δt+aΔt^2

erm … 5 out of 10 :redface:
 
  • #6
tiny-tim said:
erm … 5 out of 10 :redface:

Whoops, what I meant was;

d=v1t+.5at^2
 
  • #7
A car accelerates at 3 metres per second-squared from rest for 10s. How far does it travel?

a = 3 metres per second-squared
v1 = 0?
v2 = ?
Δd = ?
Δt = 10s

So, a is the acceleration, which is at 3 metres per second-squared, the initial velocity is 0, because the question says it starts "from rest", Δt, this will determine the interval of time used in determining the velocity, I think, not sure, and v2 and Δd is unknown. Am I right so far?
 
  • #8
Start with distance travel=average velocity × time
Here in the problem you have to find average velocity.
You are given value of acceleration which is assumed to be constant.
From this value you can find average velocity.
 
  • #9
you are right on so far
 
  • #10
A car accelerates at 3 metres per second-squared from rest for 10s. How far does it travel?

a = 3 metres per second-squared (acceleration)
v1 = 0? (initial velocity)
v2 = ? (final velocity)
Δd = ? (displacement) how far does something travel?
Δt = 10s (interval) time

So, a is the acceleration, which is at 3 metres per second-squared, the initial velocity is 0, because the question says it starts "from rest", Δt, this will determine the interval of time used in determining the velocity. v2 and Δd is unknown.
Am I right so far?
 
Last edited:
  • #11
I'm not sure what you know about Kinematics, for a start one dimension.
 
  • #12
one dimension?
 
  • #13
hi jbeannie05! :smile:

(just got up :zzz:)
jbeannie05 said:
So, a is the acceleration, which is at 3 metres per second-squared, the initial velocity is 0, because the question says it starts "from rest", Δt, this will determine the interval of time used in determining the velocity. v2 and Δd is unknown.
Am I right so far?

completely! :smile:

now look up the standard constant acceleration equations, and apply one of them :wink:
 
  • #14
A car accelerates at 3 metres per second-squared from rest for 10s. How far does it travel?

a = 3 metres per second-squared (acceleration)
v1 = 0? (initial velocity)
v2 = ? (final velocity)
Δd = ? (displacement) how far does something travel?
Δt = 10s (interval) time

So, a is the acceleration, which is at 3 metres per second-squared, the initial velocity is 0, because the question says it starts "from rest", Δt, this will determine the interval of time used in determining the velocity. v2 and Δd is unknown.

Δd = v1Δt + one half 2 squared
Δd = 0Δt + one half 3 metres per second-squared 10s 2 squared

Is that the right equation to use?
 
  • #15
jbeannie05 said:
A car accelerates at 3 metres per second-squared from rest for 10s. How far does it travel?

a = 3 metres per second-squared (acceleration)
v1 = 0? (initial velocity)
v2 = ? (final velocity)
Δd = ? (displacement) how far does something travel?
Δt = 10s (interval) time

So, a is the acceleration, which is at 3 metres per second-squared, the initial velocity is 0, because the question says it starts "from rest", Δt, this will determine the interval of time used in determining the velocity. v2 and Δd is unknown.

Δd = v1Δt + one half 2 squared
Δd = 0Δt + one half 3 metres per second-squared 10s 2 squared

Is that the right equation to use?

Δd is how far(vector) it is from the origin.
 
  • #16
jbeannie05 said:
Δd = v1Δt + one half 2 squared

Umm..what does this mean? Did you follow the link posted by tiny-tim?
 
  • #17
A car accelerates at 3 metres per second-squared from rest for 10s. How far does it travel?

a = 3 metres per second-squared (acceleration)
v1 = 0? (initial velocity)
v2 = ? (final velocity)
Δd = ? is how far(vector) it is from the origin.
Δt = 10s (interval) time

So, a is the acceleration, which is at 3 metres per second-squared, the initial velocity is 0, because the question says it starts "from rest", Δt, this will determine the interval of time used in determining the velocity. v2 and Δd is unknown.

Δd = v1Δt + ½at^​2
Δd = (0)(Δt) + ½(3m/s^​2)(10s)^​2

Is that the right equation to use?
 
  • #18
hi jbeannie05! :smile:

(try using the X2 and X2 buttons just above the Reply box :wink:)
jbeannie05 said:
Δd = v1Δt + ½at2
Δd = (0)(Δt) + ½(3m/s2)(10s)2​

Is that the right equation to use?

yup! :biggrin:
 
  • #19
Δd = (0)(Δt) + ½(3m/s2)(10s)2​

How would I calculate this on a calculator?
 
  • #20
wot? a half times 3 times 10-squared??

i did that in my head, and got 149.9999 :wink:
 
  • #21
(0)(Δt) How would I calculate this equation or what does that equation equal?
 
  • #22
What do you get when you multiply the number zero times any other number?
 
  • #23
(0)(Δt) so, the answer to that equation is 0 or zero? And I wasn't sure if (0)(Δt) is multiplying. Why is that multiplication? Is ()() always multiplication?
 
  • #24
jbeannie05 said:
Is ()() always multiplication?

ah, yes …

if there's nothing between two terms (on the same level), then it's always multiplication :wink:

(and so (0)(something) is always 0)
 
  • #25
(0)(Δt) interesting, so, if I calculate this, it is 0, (0)(anything) is always 0, interesting.

what is 0 + ½?
 
  • #26
Tiny-tim deserves the most patient helper of the year award.

How do people expect to do physics with math without the math part...

The most helpful thing would be a huge math review imo.

God Bless you tiny-tim... -Scrooge
 
  • #27
jbeannie05 said:
(0)(Δt) interesting, so, if I calculate this, it is 0, (0)(anything) is always 0, interesting.

what is 0 + ½?
You're either woefully unprepared, or you're trolling. This thread is closed.
 
  • #28
By the way, with constant acceleration, the average speed over an interval is the same as the average of the initial and final speeds.

If the car accelerated at 3 meters per second squared (NOT "3 meters squared"- you should know better) then in 10 seconds it will have gone from 0 m per second to 30 meters per second and the average of those two numbers is 15. At an average 15 meters per second, how far will it have gone in those 10 seconds?
 

1. What is the kinematics acceleration equation?

The kinematics acceleration equation is a mathematical formula used to calculate an object's acceleration. It is given by the formula a = (vf - vi) / t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time.

2. How do I use the kinematics acceleration equation?

To use the kinematics acceleration equation, you need to know the values of the final velocity, initial velocity, and time. Simply plug these values into the formula and solve for acceleration. It is important to ensure that all units are consistent (e.g. meters per second for velocity and seconds for time) to get an accurate result.

3. What is the relationship between acceleration and velocity?

Acceleration and velocity are closely related. Acceleration is the rate at which an object's velocity changes. In other words, it is the change in velocity over time. If an object has a constant acceleration, its velocity will increase or decrease at a steady rate. If the acceleration is zero, the velocity will remain constant.

4. Can the kinematics acceleration equation be used for objects with changing acceleration?

Yes, the kinematics acceleration equation can be used for objects with changing acceleration. However, in this case, the equation only gives the average acceleration over the given time interval. To calculate the acceleration at a specific instant, you would need to use calculus and take the derivative of the velocity function.

5. What are some real-life applications of the kinematics acceleration equation?

The kinematics acceleration equation is used in many real-life scenarios, such as in the design of roller coasters, cars, and other vehicles. It is also used in sports, such as calculating the acceleration of a baseball thrown by a pitcher. Additionally, it is used in physics experiments to analyze the motion of objects and in engineering to design and test structures and machines.

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