How Does Constant Acceleration Affect Distance Traveled?

AI Thread Summary
A car accelerates at 3 meters per second squared from rest for 10 seconds, leading to a calculation of distance traveled. The initial velocity is zero since the car starts from rest, and the total time of acceleration is 10 seconds. The relevant equation for distance under constant acceleration is Δd = v1Δt + ½at², which simplifies to Δd = 0 + ½(3)(10²). The final calculation shows that the car travels 150 meters in that time frame. Understanding the relationship between acceleration, time, and distance is crucial in kinematics.
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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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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)
 
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.
 
Δd=v1Δt+aΔt^2

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

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

Whoops, what I meant was;

d=v1t+.5at^2
 
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?
 
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.
 
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?
 
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