## 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?
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hi jbeannie05! welcome to pf!
 Quote by jbeannie05 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.

## Kinematics Acceleration Equation help please

Δd=v1Δt+aΔt^2

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 Quote by gbaby370 Δd=v1Δt+aΔt^2
erm … 5 out of 10

 Quote by tiny-tim 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?
 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
 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?
 I'm not sure what you know about Kinematics, for a start one dimension.
 one dimension?

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hi jbeannie05!

(just got up )
 Quote by jbeannie05 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?

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

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 Quote by jbeannie05 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?