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Tentothe

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**[SOLVED] Uniformly Accelerated Linear Motion**

Hello,

I am attempting what looks to be a fairly easy problem involving uniform acceleration and linear motion, but my answers just aren't coming out right and I'm unable to see where I'm going wrong. Any help is appreciated.

## Homework Statement

A teenager has a car that accelerates at 3.00 m/s[tex]^{2}[/tex] and decelerates at -4.50 m/s[tex]^{2}[/tex]. On a trip to the store, he accelerates from rest to 15.0 m/s, drives at a constant speed for 6.00 s, and then comes to a momentary stop at the corner. He then accelerates to 18.0 m/s, drives at a constant speed for 15.0 s, and decelerates for 2.67 s. Then he continues for 4.00 s at this speed and comes to a stop.

(a) How long does the trip take?

(b) How far has he traveled?

## Homework Equations

(i) [tex]\Delta[/tex]X = [tex]\frac{v^{2}_{f} - v^{2}_{i}}{2a}[/tex]

(ii) [tex]\Delta[/tex]X = [tex]v_{i}t[/tex] + [tex]\frac{1}{2}at^{2}[/tex]

(iii) [tex]v = v_{i} + at[/tex]

## The Attempt at a Solution

I've broken the problem down into the 6 parts of the trip and calculated both the time and distance over each and then added them up for the total.

__(a) Calculating time taken__

*Part I*: Accelerates from rest to 15.0 m/s, so:

[tex]v_{i}[/tex] = 0 m/s, [tex]v_{f}[/tex] = 15.0 m/s, a = 3.00 m/s[tex]^{2}[/tex]

Solving equation (iii) for t and plugging in values gives:

t = (15.0 - 0)/3.00 = 5.00 s.

*Part II*: Constant speed for 6.00 s, so no calculation needed; t = 6.00 s.

*Part III*: After momentary stop, accelerates from rest to 18.0 m/s, so same as part I.

[tex]v_{i}[/tex] = 0 m/s, [tex]v_{f}[/tex] = 18.0 m/s, a = 3.00 m/s[tex]^{2}[/tex]

t = (18.0 - 0)/3.00 = 6.00 s.

*Part IV*: Constant speed for 15.0 s, so again no calculation needed; t = 15.0 s.

*Part V*: Decelerates for 2.67 s; time given, no calculation needed; t = 2.67 s.

*Part VI*: Constant speed for 4.00 s, no calculation needed; t = 4.00 s.

Adding up time taken for all parts: 5.00 s + 6.00 s + 6.00 s + 15.0 s + 2.67 s + 4.00 s =

**.**

__38.67 s____(b) Calculating distance traveled__

*Part I*: Using equation (i) with [tex]x_{i}[/tex] = 0, [tex]v_{f}[/tex] = 15.0 m/s, [tex]v_{i}[/tex] = 0 m/s, a = 3.00 m/s[tex]_{2}[/tex],

X = (15.0)[tex]^{2}[/tex]/2(3.00) = 37.5 m

*Part II*: Constant speed of 15.0 m/s for 6.00 s, so X = 15.0 * 6.00 = 90.0 m.

*Part III*: Same as part I after stop, so using equation (i) with [tex]x_{i}[/tex] = 0, [tex]v_{f}[/tex] = 18.0 m/s, [tex]v_{i}[/tex] = 0 m/s, a = 3.00 m/s[tex]^{2}[/tex],

X = (18.0)[tex]^{2}[/tex]/2(3.00) = 54.0 m

*Part IV*: Constant speed of 18.0 m/s for 15.0 s, so X = 18.0 * 15.0 = 270 m.

*Part V*: Decelerates for 2.67 s, so using equation (ii) with [tex]v_{i}[/tex] = 18.0 m/s, t = 2.67 s, a = -4.50 m/s[tex]^{2}[/tex],

X = (18.0)(2.67) + (0.5)(-4.50)(2.67)[tex]^{2}[/tex] = 32.0 m

*Part VI*: Constant speed for 4.00 s, so first need to find [tex]v_{f}[/tex] after deceleration in part V by using equation (iii) with [tex]v^{i}[/tex] = 18.0 m/s, a = -4.50 m/s[tex]^{2}[/tex], and t = 2.67 s.

v = (18.0) + (-4.50)(2.67) = 5.99 m/s

Travels at constant speed of 5.99 m/s for 4.00 s, so 5.99 * 4.00 = 23.9 m.

Adding up all of the distances gives total: 37.5 m + 90.0 m + 54.0 m + 270 m + 32.0 m + 23.9 m =

**.**

__507.4 m__I must be missing something that's glaringly obvious here. I thought perhaps the undefined "momentary stop" might be skewing my results for part (a), but more likely I've overlooked something. Thanks in advance to whoever can point out where I've gone wrong.