# Uniformly Accelerated Linear Motion

• Tentothe
In summary, the problem involves a teenager driving a car with uniform acceleration and deceleration. The total time and distance of the trip is calculated by breaking it down into 6 parts and adding up the individual times and distances. After realizing a missing calculation, the correct answer is obtained.
Tentothe
[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$$^{2}$$ and decelerates at -4.50 m/s$$^{2}$$. 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) $$\Delta$$X = $$\frac{v^{2}_{f} - v^{2}_{i}}{2a}$$

(ii) $$\Delta$$X = $$v_{i}t$$ + $$\frac{1}{2}at^{2}$$

(iii) $$v = v_{i} + at$$

## 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:
$$v_{i}$$ = 0 m/s, $$v_{f}$$ = 15.0 m/s, a = 3.00 m/s$$^{2}$$​

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.
$$v_{i}$$ = 0 m/s, $$v_{f}$$ = 18.0 m/s, a = 3.00 m/s$$^{2}$$​
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 $$x_{i}$$ = 0, $$v_{f}$$ = 15.0 m/s, $$v_{i}$$ = 0 m/s, a = 3.00 m/s$$_{2}$$,
X = (15.0)$$^{2}$$/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 $$x_{i}$$ = 0, $$v_{f}$$ = 18.0 m/s, $$v_{i}$$ = 0 m/s, a = 3.00 m/s$$^{2}$$,
X = (18.0)$$^{2}$$/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 $$v_{i}$$ = 18.0 m/s, t = 2.67 s, a = -4.50 m/s$$^{2}$$,
X = (18.0)(2.67) + (0.5)(-4.50)(2.67)$$^{2}$$ = 32.0 m​

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

and then comes to a momentary stop at the corner.

I wonder how he managed to come to that momentary stop, without decelerating ?

>>Then he continues for 4.00 s at this speed and comes to a stop.

I think there will be a Part VII for calculating the distance traveled while decelerating.

x=(0^2-5.99^2)/(2*(-4.50))How do you type the square ? I don't know how to get that square so i use ^

I think there will be a Part VII for calculating the distance traveled while decelerating.

Thanks. I knew I was excluding something. I just factored in those 2 stops and got the answer.

How do you type the square ? I don't know how to get that square so i use ^

I used the Latex code this forum has. There is a reference for all of the codes on the post message page. You can click on the sigma icon and then "Subscript and Superscript," then the picture showing a superscript and type a '2' between the brackets. The code is something like x[ tex]^{2}[ /tex] without the spaces to get x$$^{2}$$. A lot faster to just type ^ though.

## 1. What is uniformly accelerated linear motion?

Uniformly accelerated linear motion is a type of motion in which an object travels in a straight line with a constant acceleration. This means that the object's velocity increases or decreases by the same amount over equal intervals of time.

## 2. How is uniformly accelerated linear motion different from uniformly linear motion?

Uniformly accelerated linear motion is different from uniformly linear motion in that the former involves a constant acceleration, while the latter involves a constant velocity. In other words, in uniformly accelerated linear motion, the object is either speeding up or slowing down, whereas in uniformly linear motion, the object maintains the same speed throughout its motion.

## 3. How is acceleration calculated in uniformly accelerated linear motion?

The formula for calculating acceleration in uniformly accelerated linear motion is a = (vf - vi) / t, where a is acceleration, vf is the final velocity, vi is the initial velocity, and t is the time interval. The unit for acceleration is meters per second squared (m/s^2).

## 4. What is the difference between positive and negative acceleration in uniformly accelerated linear motion?

In uniformly accelerated linear motion, positive acceleration refers to an increase in speed, while negative acceleration (also known as deceleration) refers to a decrease in speed. This is reflected in the equations for acceleration, where a positive value indicates speeding up and a negative value indicates slowing down.

## 5. Can an object have both a constant velocity and a constant acceleration in uniformly accelerated linear motion?

No, an object cannot have both a constant velocity and a constant acceleration in uniformly accelerated linear motion. This is because a constant acceleration requires a change in velocity over time, while a constant velocity means there is no change in velocity over time.

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