# Kinematics Acceleration Equation help please

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?

tiny-tim
Homework Helper
welcome to pf!

hi jbeannie05! welcome to pf!
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?

(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

tiny-tim
Homework Helper
Δd=v1Δt+aΔt^2

erm … 5 out of 10

erm … 5 out of 10

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?

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

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:
I'm not sure what you know about Kinematics, for a start one dimension.

one dimension?

tiny-tim
Homework Helper
hi jbeannie05!

(just got up :zzz:)
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!

now look up the standard constant acceleration equations, and apply one of them

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?

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.

Δd = v1Δt + one half 2 squared

Umm..what does this mean? Did you follow the link posted by tiny-tim?

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?

tiny-tim
Homework Helper
hi jbeannie05!

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

Is that the right equation to use?

yup!

Δd = (0)(Δt) + ½(3m/s2)(10s)2​

How would I calculate this on a calculator?

tiny-tim
Homework Helper
wot? a half times 3 times 10-squared??

i did that in my head, and got 149.9999

(0)(Δt) How would I calculate this equation or what does that equation equal?

TSny
Homework Helper
Gold Member
What do you get when you multiply the number zero times any other number?

(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?

tiny-tim
Homework Helper
Is ()() always multiplication?

ah, yes …

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

(and so (0)(something) is always 0)

(0)(Δt) interesting, so, if I calculate this, it is 0, (0)(anything) is always 0, interesting.

what is 0 + ½?