Please explain this Kinematic Problem.

[vpa021]

How far do you travel if, starting from rest, you accelerate at 1.5 m/s2 for
4.0 s, then coast for 8.0 seconds, then decelerate to a standstill in 4.0 s?

I am totally confused about the fact you accelerate and then coast, and then slow down. We have 3-t's in this problem and don't know what formula to use.

 

[wukunlin]

you probably want to divide the problem into these three stages. I find V-T diagrams helpful in these problems

 

[vpa021]

you probably want to divide the problem into these three stages. I find V-T diagrams helpful in these problems

What are V-T diagrams?

 

[iRaid]

Think of it like this:
The first part, you are accelerating at 1.5m/s (per second), therefore, after 4s you will be at v=??. Then the next part, you are coasting, therefore, not accelerating.

 

[vpa021]

Think of it like this:
The first part, you are accelerating at 1.5m/s (per second), therefore, after 4s you will be at v=??. Then the next part, you are coasting, therefore, not accelerating.

So basically, I should just brake the question up into 3 steps?

 

[wukunlin]

sorry I mean velocity vs. Time diagrams, with area under the graph representing your displacement. You will need to use kinematics equation to find your peak velocity

 

[vpa021]

OK. I will try to work this out the way you told me to.

 

[Sefrez]

Just look at it as three separate problems. First you travel distance when accelerating (constant acceleration,) then when you coast (constant velocity,) and then when you decelerate.

Spoiler

The first you are given that you start from rest and accelerate at 1.5m/s^2 for 4 seconds. Using the derived expression for displacement in terms of what you are given:
ΔD = v_i*t + 0.5*a*t^2 (v_i = velocity initial = 0) = 0.5*1.5m/s^2*(4s)^2 = 12m

Next you are given that you coast for 8.0s. This is simply given by:
ΔD = v_f*t (v_f = velocity final of last part)

Where v_f = v_i + at = 0 + 1.5m/s^2*4.0s = 6.0m/s

So
ΔD = 6.0m/s*8.0s = 48m

For the last part you are given that you come to a stop in 4.0s. A short cut here is to pickout that this is the same amount of time that was taken to reach v_f. With constant deceleration, then, you know the same distance is taken as was when starting out, 12m.

But to find it anyway, you have current velocity as 6.0m/s. The acceleration (in other direction) required for that to reach zero in 4.0s can be taken from v_f = v_i + at where v_f = 0m/s, v_i = 6.0m/s, and t = 4.0s. So 0 = 6.0m/s + a*4.0s => a = -1.5m/s^2. Which is expected as it is the same in magnitude as the beginning acceleration. So to get displacement:
ΔD = v_i*t + 0.5at^2 = 6m/s*4.0s - 0.5*1.5m/s^2*(4s) = 12m

Which is as said, the same as the displacement for part one. So total displacement is:
D = 12m + 48m + 12m = 72m

 

[vpa021]

thank you sefrez.

 

[Sefrez]

I did that because at first you are accelerating for 4 seconds. The next part you are coasting for 8 seconds. Your velocity is constant here.

Edit, I see you edited your post. You must have figured out. If I had waited a few more seconds lol.

 

[vpa021]

I did that because at first you are accelerating for 4 seconds. The next part you are coasting for 8 seconds. Your velocity is constant here.

Edit, I see you edited your post. You must have figured out. If I had waited a few more seconds lol.

Yes... lol i figured it out :).
You can be sure you will be seeing me very often on this forum.

 

[PeterO]

What are V-T diagrams?

V-T graphs represent, what historians would call, the "primary source", (the original information) from which the 5 equations of motion are derived. Well four are derived from a V-T graph and the other is a mathematical manipulation of a couple of them.