Motion problems

[IbelieveIAm]

Hi,

 

I wonder if someone could help me with two motion problems.

 

The first one is:

 

A car can accelerate at 1.9m.s-2. It is initially travelling at 19m.s-1 when it passes through an intersection and it then accelerates for the next 85m. What is the final velocity of the car?

 

And the second one is:

 

A car is travelling at 95km/h for 130km, and then slows down to 65km/h. It takes 3 hrs and 20 mins. How far has the car travelled?

 

If someone can help, I would be very grateful!!

 

God bless.

[swansont]

Hi,

 

I wonder if someone could help me with two motion problems.

 

The first one is:

 

A car can accelerate at 1.9m.s-2. It is initially travelling at 19m.s-1 when it passes through an intersection and it then accelerates for the next 85m. What is the final velocity of the car?

 

And the second one is:

 

A car is travelling at 95km/h for 130km, and then slows down to 65km/h. It takes 3 hrs and 20 mins. How far has the car travelled?

 

If someone can help, I would be very grateful!!

 

God bless.

 

 

What equations can you use here?

[spunnery]

Hi,

 

I wonder if someone could help me with two motion problems.

 

The first one is:

 

A car can accelerate at 1.9m.s-2. It is initially travelling at 19m.s-1 when it passes through an intersection and it then accelerates for the next 85m. What is the final velocity of the car?

 

And the second one is:

 

A car is travelling at 95km/h for 130km, and then slows down to 65km/h. It takes 3 hrs and 20 mins. How far has the car travelled?

 

If someone can help, I would be very grateful!!

 

God bless.

The first answer is 31.709 m /sec.

Second answer is 257.719 km

[ydoaPs]

The first answer is 31.709 m /sec. first find time = 6.68856 seconds 85 m / t^2 = 1.9 m / sec^2;Final velocity wll be (85/6.68856) + 19 (initial velocity).

Second answer is 257.719 km - (retardation time considered 0)

We're not supposed to give homework answers. We can help them do it themselves, but we are not to do their homework for them. You can give equations and hints, but not answers.

[spunnery]

We're not supposed to give homework answers. We can help them do it themselves, but we are not to do their homework for them. You can give equations and hints, but not answers.

Sorry I was not aware about that.really sorry

[dstebbins]

Hi,

 

I wonder if someone could help me with two motion problems.

 

The first one is:

 

A car can accelerate at 1.9m.s-2. It is initially travelling at 19m.s-1 when it passes through an intersection and it then accelerates for the next 85m. What is the final velocity of the car?

Well, start by using the distance formula, which is d=(at^2)/2. Once you find the time it takes to travel these 85m at a constant acceleration of 1.9m/s/s, then use the acceleration equation, which is a=(v2-v1)/t, and solve for v2.

 

And the second one is:

 

A car is travelling at 95km/h for 130km, and then slows down to 65km/h. It takes 3 hrs and 20 mins. How far has the car travelled?

 

If someone can help, I would be very grateful!!

 

God bless.

 

Follow these steps.

 

1. Convert all those measurements to base units. Convert kilometers to meters and hours and minutes to seconds. It's not mandatory, but it makes it easier (a LOT easier, actually).

 

2. Find out how long it took the car to travel the first 130km. Subtract that time from 3hrs 20mins. That's how long it takes to travel the remainder.

 

3. after that, use the velocity formula, which is d=vt.

[IbelieveIAm]

Thank you to everyone who posted help!! Thanks for taking the time!

[dstebbins]

Thank you to everyone who posted help!! Thanks for taking the time!

 

no problem. It's what this board is here for.

 

Besides, it makes me feel smart when I can help people out on their homework.

[Jacques]

Well, start by using the distance formula, which is d=(at^2)/2. Once you find the time it takes to travel these 85m at a constant acceleration of 1.9m/s/s, then use the acceleration equation, which is a=(v2-v1)/t, and solve for v2.

Distance formula you used is for initial velocity = 0...

but the initial velocity is 19 m.s-1

d=(at^2)/2 + vt

[dstebbins]

^^^And how would you solve for t in that equation? I'm lost as hell.

[Jacques]

You have a quadratic equation

[math]d=at^2/2 + vt[/math]

[math]0=at^2/2 + vt -d[/math]

remember the good old solution [math]-b +- \sqrt (b^2 -4ac) / 2a[/math]

 

Just plug the good value in that Quadratic equation solver

and you will get the time the acceleration took place and from there I am sure you can get it

[dstebbins]

I see. One more thing.

 

The first answer is 31.709 m /sec.

Second answer is 257.719 km

 

Your're wrong, not because you came up with the wrong number, but because you didn't express your answer to the proper number of significant digits.

[foodchain]

I see. One more thing.

 

 

 

Your're wrong, not because you came up with the wrong number, but because you didn't express your answer to the proper number of significant digits.

 

what is that, four or something:confused: Sometimes the intensity of math simply turns me away, then I realize though that such exists for good reasons overall, or at least positive reasons, or I would at least hope so.

[dstebbins]

Ah, then allow your's truly to explain.

 

First off, let's get some things straight right off the bat: 1) There is no such thing as 100% accuracy, simply because no matter how many decimal places you round to, there's always more that you can round to. 2) A calculation is only as accurate as its least accurate measurement, just the same as a society is only as successful as its least fortunate members. 3) Significant digits are used to measure accuracy, and because a calculation is only as accurate as its least accurate measurement, your answer must be in the same number of significant digits as the least accurate measurement (aka the measurement with the least number of SDs.)

 

But that all boils down to the question: What is an SD? Well, here's a list of rules for a number to be an SD. If it meets just ONE of these rules, it's an SD.

 

1. All nonzero digits.

2. All final zeroes after a decimal point

3. All zeroes between other significant digits.

4. All zeroes with a nonzero to the left of them and a decimal to the right.

 

So let's put this to the test.

 

1.0 would have two SDs, since 1 is a nonzero digit and the 0 is a final zero after a decimal point. 10.0 would have three, since the one is nonzero, the second zero is final after a decimal, and the first zero is between two SDs. 10 would have only one since the zero is final, but not after a decimal. 10. (read as "ten point") would have two since rule #4 would apply. .01 would have one SD, since #4 would apply if the 1 and decimal were switched, and finally, .010 would have two, since it's the same as the last one, except there's a final zero after a decimal.

 

Any questions?