# Finding Distance Given Velocity and Time

• Simon777
I think you should think of this as the displacement and not absolute distance.You're right, I slacked off with that for this problem since it all occurs in one direction and went with magnitude alone for position, but that's inconsistent since I used a vector quantity for change in position. I'll keep this in mind for future problems, thank you.No problem! It's always good to keep track of the units and whether we are dealing with scalars or vectors. It can make a big difference in the final answer.Glad I could help! And don't worry, physics can be challenging but it's all about practice and getting familiar with the concepts. Keep at it and you'll be an expert in no time
Simon777

## Homework Statement

A car accelerates uniformly and reaches a speed of 4.40 m/s in 8.00 seconds. Calculate the distance traveled by the car from a time of 1.40 to 4.70 seconds.

## Homework Equations

x = xo + vo t + ½ a t^2

(v-vo)/t=a

## The Attempt at a Solution

(v-vo)/t=a
(4.40 m/s - 0 m/s)/8.00s=a
.550 m/s^2=a

x = xo + vo t + ½ a t^2
x= 0 + 0(3.30s) + 1/2 (.550 m/s^2) (3.30s)^2
x= 3.00 meters

My homework is on LON-CAPA, an online program and when I enter 3.00 meters it states the answer is incorrect. I've been working on this problem for 2 hours and can't come up with any other answer.

Simon777 said:
x= 0 + 0(3.30s) + 1/2 (.550 m/s^2) (3.30s)^2
x= 3.00 meters

My homework is on LON-CAPA, an online program and when I enter 3.00 meters it states the answer is incorrect. I've been working on this problem for 2 hours and can't come up with any other answer.
Is the velocity really zero at t=1.40 seconds?

Simon777 said:

## Homework Statement

A car accelerates uniformly and reaches a speed of 4.40 m/s in 8.00 seconds. Calculate the distance traveled by the car from a time of 1.40 to 4.70 seconds.

## Homework Equations

x = xo + vo t + ½ a t^2

(v-vo)/t=a

## The Attempt at a Solution

(v-vo)/t=a
(4.40 m/s - 0 m/s)/8.00s=a
.550 m/s^2=a

x = xo + vo t + ½ a t^2
x= 0 + 0(3.30s) + 1/2 (.550 m/s^2) (3.30s)^2
x= 3.00 meters

My homework is on LON-CAPA, an online program and when I enter 3.00 meters it states the answer is incorrect. I've been working on this problem for 2 hours and can't come up with any other answer.

Looks like you have calculated where the car will be at time 3.30s to me??

That is not what you wanted - not what you thought you were finding.

My logic for that is if acceleration is constant, the rate of change of velocity will be constant so you'd cover the same distance from time 0s to 3.30s as you would 1.40s to 4.70s.

Is at least the acceleration I found correct?

Simon777 said:
My logic for that is if acceleration is constant, the rate of change of velocity will be constant so you'd cover the same distance from time 0s to 3.30s as you would 1.40s to 4.70s.
That would be true if the velocity was constant, which sadly it isn't.
Simon777 said:
Is at least the acceleration I found correct?
Yes, your acceleration is indeed correct.

So, in order to correct your formulae, you need to work out the velocity at t=1.40 seconds.

Hootenanny said:
That would be true if the velocity was constant, which sadly it isn't.

Yes, your acceleration is indeed correct.

So, in order to correct your formulae, you need to work out the velocity at t=1.40 seconds.

I guess I mistakenly thought it was constant because it increases at a constant rate. So in physics, does constant only mean a value stays the same during a period of time?

I'll rework this and post what I get soon. Thank you so much for the help.

I think you should think of this as the displacement and not absolute distance.

Simon777 said:
I guess I mistakenly thought it was constant because it increases at a constant rate. So in physics, does constant only mean a value stays the same during a period of time?
In a manner of speaking, yes. In Physics, constant means unchanging. Otherwise, how would we differentiate between something that remained constant, or changed at a constant rate
Simon777 said:
I'll rework this and post what I get soon. Thank you so much for the help.
No problem!

v(1.40s)=.770 m/s

x = xo + vo t + ½ a t^2
x= 0 + (.770 m/s) (3.30s) + 1/2 (.55 m/s^2) (3.30s)
x= 5.54 m

Simon777 said:
v(1.40s)=.770 m/s

x = xo + vo t + ½ a t^2
x= 0 + (.770 m/s) (3.30s) + 1/2 (.55 m/s^2) (3.30s)
x= 5.54 m
Looks okay to me

(Although there is a small typo in the second line - it should be (3.30s)^2)

WatermelonPig said:
I think you should think of this as the displacement and not absolute distance.

You're right, I slacked off with that for this problem since it all occurs in one direction and went with magnitude alone for position, but that's inconsistent since I used a vector quantity for change in position. I'll keep this in mind for future problems, thank you.

Simon777 said:
I guess I mistakenly thought it was constant because it increases at a constant rate. So in physics, does constant only mean a value stays the same during a period of time?
Not only in Physics.

You can sit in a parked car for 10 minutes and your position, velocity and acceleration will all be constant, ie zero.

You can sit in a car for 10 minutes that is being driven at 20 miles an hour and your speed and acceleration will be constant [one of them will be zero] but your position will certainly change.

You can sit in a car for 30 seconds that is easing away from traffic lights accelerating at a constant 2 miles per hour, per second.
All the while your acceleration will be constant, but your speed sure is changing - you will be doing 60 mph at the end of the 30 seconds!
And your position will be changing as well.

EDIT: I used miles per hour since the speedometer in a car is not calibrated to metres per second. Change miles to kilometres if you are from a metric country.

Last edited:
Hootenanny said:
Looks okay to me

(Although there is a small typo in the second line - it should be (3.30s)^2)

Whoops, nice catch. I see where I went wrong in my logic now. I made a graph of velocity over time and the area under the line from 0s to 3.3s is much smaller than the area under the line from 1.40s to 4.70s. The only way to make what I originally did work is if, like you said, velocity was constant, making both these areas under the line equivalent.

Thanks again for the help, everyone on this forum is always so kind and I really appreciate people like you who use their free time to help those struggling with this subject.

## What is the formula for finding distance given velocity and time?

The formula for finding distance given velocity and time is d = vt, where d is the distance, v is the velocity, and t is the time.

## How do you calculate distance when given velocity and time?

To calculate distance when given velocity and time, you simply multiply the velocity by the time. This will give you the distance traveled during that time period.

## What units should be used for velocity and time in the distance formula?

The units for velocity should be in distance per time (ex: meters per second, miles per hour), and the unit for time should be in seconds, minutes, or hours. It is important to make sure the units are consistent in order to get an accurate answer.

## Can the distance formula be used for any type of motion?

Yes, the distance formula can be used for any type of motion as long as the velocity and time are known. It can be used for linear motion, circular motion, and even accelerated motion.

## What is the difference between speed and velocity in the distance formula?

Speed is a scalar quantity that only measures how fast an object is moving, while velocity is a vector quantity that measures both the speed and direction of an object's motion. In the distance formula, the velocity is used because it accounts for the direction of motion, not just the speed.

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