Kinematics Acceleration Equation help please

jbeannie05

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

hi jbeannie05! welcome to pf!
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)

oli4

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.

gbaby370

Δd=v1Δt+aΔt^2

This should answer your question.

tiny-tim

erm … 5 out of 10

gbaby370

Whoops, what I meant was;

d=v1t+.5at^2

jbeannie05

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?

azizlwl

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.

sckeen1988

you are right on so far

jbeannie05

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?

azizlwl

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

jbeannie05

one dimension?

tiny-tim

hi jbeannie05!

(just got up )
completely!

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

jbeannie05

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?

azizlwl

Δd is how far(vector) it is from the origin.

Pranav-Arora

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

jbeannie05

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?