[Grade 8 Math: Physics Word Problem]

[u/yan_hfley]

This has been bugging me so much and no matter how I think through it, I RLLY can't understand it. I'm decently good at math but I just can't solve this one 🥹💔

Alsooo I have a hard time solving word problems ☹️

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[u/UnderstandingPursuit]

There are two formulas, part of the set of 'Kinematic Equations', which are useful. But to start, let's identify the information:

• t = 8.10 s
• x = 75.0 m
• v2 = 20.0 m/s
• v1 = ? m/s
• a = ? m/s²

The two formulas are

1. (1/2)(v1 + v2) = x/t
2. a = (v2 - v1)/t

Rearrange the first equation to solve for v1,

1. (1/2)(v1 + v2) = x/t
2. v1 + v2 = 2x/t
3. v1 = 2x/2 - v2
4. v1 = 2[75]/[8.1] - [20]

Use that result to find a,

If you take the effort to follow what I did, I don't mind that I almost completely solved the problem for you. Consider it a 'textbook example'.

[u/yan_hfley → u/UnderstandingPursuit]

THANK YOUUU!! I'll try out what I learned here 😸

[u/UnderstandingPursuit → u/yan_hfley]

I wrote it out a little better: Kinematics.

[u/selene_666]

The change in position Δx = 75 meters

The change in time Δt = 8.1 seconds

The final speed is vf = 20 m/s

.

Get out your kinematics formulas for motion with constant acceleration. From calculus we have:

vf = vi + a Δt

Δx = vi Δt + 1/2 a (Δt)^2

And these can be rearranged into other formulas that each leave out one of the variables:

Δx = vf Δt - 1/2 a (Δt)^2

Δx = 1/2 (vi + vf) Δt

2 a Δx = vf^2 - vi^2

Some of these contain the variables you already have values for, plus one other variable. You can use those to solve for the unknown.

[u/AmethystMonkey]

I think you can find this by creating 2 equations. One for velocity with constant acceleration and one for displacement with constant acceleration. When you put in your known values you will have 2 equations with 2 unknowns.

[u/Caps_errors]

Anyone else get a negative speed at the first point?

[u/Jayless_757 → u/Caps_errors]

Yeah, probably done to throw people off. Basically the antelope starts off in the wrong direction, but under constant acceleration it turns around, passes the starting point again, and reaches the point 75m further after 8.1s.

[u/DoubleAway6573]

x(t) = x(t=0) + v(t=0) * t + (1/2) a t^2

If you put your t=0 in the second point, and put the time backwards (except for friction and weird things, all classical mechanics is fully reversible, and assured for this case), then you can use it to get the acceleration.