How Do You Calculate Velocity and Distance in Physics Problems?

[terpsgirl]

I've been trying to figure them out but can't get through about 6 questions, these are the other two..

An object is observed to fall from a bridge, striking the water below 2.50 s later. (a) With what velocity did it strike the water? (b) What was its average velocity during the fall? (c) How high is the bridge?

A ferryboat requires 45.0 min to travel 15.0 km across a bay. (a) what was the avg speed in m/s? (b) At this speed, how much time would be required for the boat to take a 420 km trip up the river?
THX

 

[Zorodius]

I've been trying to figure them out but can't get through about 6 questions, these are the other two..

An object is observed to fall from a bridge, striking the water below 2.50 s later. (a) With what velocity did it strike the water? (b) What was its average velocity during the fall? (c) How high is the bridge?

A ferryboat requires 45.0 min to travel 15.0 km across a bay. (a) what was the avg speed in m/s? (b) At this speed, how much time would be required for the boat to take a 420 km trip up the river?
THX

For the bridge question, you know that the current velocity is given by $v = v_i + at$. You know that $v_i = 0, a = 9.8 \frac{m}{s^2}$, and $t = 2.50 s$. Plugging those numbers into the first equation, what number comes out?

What about the distance it travelled? You know that you can relate distance, initial velocity, acceleration, and time with $x = v_i t + \frac{1}{2}at^2$, where x is the displacement. Using the numbers you already have from the previous part, what do you get for x? This is the distance the object has fallen - the same thing as the "height of the bridge".

Average velocity is the ratio of displacement to time, $v_{avg} = \frac{\Delta x}{\Delta t}$. You know the displacement from your recent work, and you know the time is 2.50 s, because that's given to you in the problem. What do you get when you divide the displacement by 2.50 s? That's the average velocity.

For the ferryboat question, you can find the average velocity just like you did in the bridge question. Take the displacement (15.0 km) and divide it by the time (45.0 min). That's the average velocity. But it will be in kilometers per minute, and the problem wants you to give the average velocity in meters per second. No problem! Just multiply by a conversion factor so that the units you don't want go away, and the units you do want are all that's left. Remember, a conversion factor is equal to one, you can multiply anything you want by a conversion factor and you won't change its value.

$$\frac {15.0 km}{45.0 min} \cdot \frac {1000 m}{1 km} \cdot \frac {1 min}{60 s} = 5.56 m/s$$

If you think of the units, like "km" and "min", as being just like variables, you can see that on the left hand side all the "km"s cancel out, as do all the "min"s, and you're just left with regular meters and seconds.

To answer part (b), let's first get 420 km into the same units as the speed we know (meters):
$$420 km \cdot \frac {1000 m}{1 km} = 420000 m$$

So, what we're really asking is, how long does the boat need to go at 5.56 m/s before it has moved 420,000 m?

We know that, with a constant velocity, $x = vt$, displacement is the product of velocity and time. If we solve that for t, we get:
$$\frac {x}{v} = t$$

We know x (420,000 m) and v (5.56 m/s), so when we divide those out, what do we get?

$$\frac {420000 m}{5.56 m/s} = 75600 s$$

It will take 75,600 seconds for the boat to go that distance. That's a pretty unwieldy number, though, so let's convert it into hours by dividing by 3600. 75,600 seconds is 21 hours.

 

[M98Ranger]

Hi there,

I understand that you are stuck on these questions and have been trying to figure them out. I can see that you have two different problems, one involving an object falling from a bridge and the other involving a ferryboat traveling across a bay.

For the first problem, we can use the equation d = v*t (distance = velocity * time) to solve for the velocity of the object. We know that the time is 2.50 seconds and the distance is the height of the bridge (which we will solve for in part c). So, the equation becomes d = v*2.50. To solve for v, we need to rearrange the equation to v = d/2.50. This means that the velocity of the object when it strikes the water is equal to the height of the bridge divided by 2.50.

For part b, we need to find the average velocity during the fall. This can be calculated by taking the total distance (height of the bridge) and dividing it by the total time (2.50 seconds). So, the average velocity during the fall is the height of the bridge divided by 2.50 seconds.

Moving on to part c, we need to find the height of the bridge. We can use the same equation, d = v*t, but we will use the average velocity from part b and the total time of 2.50 seconds. This will give us the height of the bridge.

For the second problem, we are given the time and distance of the ferryboat traveling across the bay. To find the average speed in m/s, we first need to convert the time to seconds (45 minutes = 2700 seconds) and the distance to meters (15 kilometers = 15,000 meters). Then, we can use the equation v = d/t to solve for the average speed in m/s.

For part b, we are given the distance of 420 kilometers and we now know the average speed in m/s. We can use the same equation, v = d/t, to solve for the time. Just remember to convert the distance to meters and the speed to kilometers per second before plugging them into the equation.

I hope this helps you with these questions. Good luck!