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SUVAT equation help?

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1. I'm having trouble figuring out how to get the answer to this word problem:

A car leaves a set of traffic lights from stationary and accelerates at a constant rate of 6m/s^2. A lorry passes the traffic lights just as they go green. It is travelling at a constant speed of 18m/s. How many seconds after it leaves the traffic lights does the car overtake the lorry?
2. (Original post by vsvpop)
I'm having trouble figuring out how to get the answer to this word problem:

A car leaves a set of traffic lights from stationary and accelerates at a constant rate of 6m/s^2. A lorry passes the traffic lights just as they go green. It is travelling at a constant speed of 18m/s. How many seconds after it leaves the traffic lights does the car overtake the lorry?
You need to work out the time at which the distances travelled are equal
3. Eww mechanics eww SUVAT. Just nope - hopefully someone here can help you out but wish you luck for mechanics x
4. Okay so, consider the car first. s=x u=0 a=6 t=T
lorry: s=x u=18 a=0 t=T

use s=ut + 0.5at^2 to equate distances and solve for T
5. Look at the Suvat equations, you need two, one for the car and one for the lorry.

For each problem, you want to find an equation that only has one unknown in it, in this case, time.

So, you have, for the car

s = 1/2 a t^2

and for the lorry

s = v t

so that

v t = 1/2 a t^2; do some algebra and solve for t;

a good way to do these also, for practice, is to graph it:
the car is the curved line, the lorry is the straight line, they pass when the lines touch (x axis is time, y axis is distance, in metres)

https://www.desmos.com/calculator/kolzu1qxs7
6. pippa's response is good but what you need to learn is to identify the information they've given you.

If something starts "from stationary", does that tell you something about its initial velocity? Of course it does. A stationary object isn't moving, so the magnitude of its velocity is zero.

u = 0

"and accelerates at a constant rate..."

It's basically just giving you the acceleration here.

a = 6

"The lorry passes the traffic light just as they go green"

It's telling you that they're starting from the same instant, because they'd be violating traffic laws to start from an earlier point. It also means that you need to use simultaneous equations.

It is travelling at a constant speed, meaning it's not accelerating (Newton's First Law)

u = 18
a = 0

The time and displacement of both will be the same when they meet (by definition, the point of overtaking is the time where the two vehicles meet ie have equal displacements).

The two unknowns are displacement (s) and time (t) using the equations pippa's given you. The equations used can be figured out by knowing what variables you have (u, a, s, t) and looking at the formulae sheet.

If you do them simultaneously by substitution, it cancels the displacement variable and gives you something to solve for time.

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Updated: July 13, 2016
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