Calculate Average Velocity for 1.5hr Bike Ride

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Homework Help Overview

The discussion revolves around calculating average velocity from a bike ride and understanding the implications of acceleration in a physics context. The original poster presents a scenario involving two segments of a bike ride and seeks to determine the average velocity. Additionally, there is a subsequent question regarding the safe stopping distance for a car hitting a barrier, incorporating concepts of acceleration and initial velocity.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the calculation of average velocity by considering total distance and total time. Some question the necessity of unit conversion, while others express confusion about the application of acceleration in a different context involving car safety.

Discussion Status

The discussion is active, with participants providing insights and corrections to each other's reasoning. There is a mix of attempts to clarify concepts and calculations, with some participants questioning assumptions and definitions related to acceleration and velocity.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the depth of guidance provided. There is an emphasis on understanding the relationships between distance, time, velocity, and acceleration without providing direct solutions.

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You ride your bike for 1.5 hour at an average velocity of 10km/hr, then for 30 min at 15km/hr. What is your average velocity?

so since average velocity = delta D/delta T

i got 7500 m / 7200 secs = 1.04 m/s.. is that correct?
 
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1.5 hours at 10 km/hr would be 15 km, right?

0.5 hours at 15 km/hr would be 7.5 km, right?

Total distance = 15 km + 7.5 km = 22.5 km, total time = 2 hours.

Total distance / total time = 22.5 km / 2 hours = ...

There's no reason at all to convert anything to meters or seconds. Leave it all in km/hr.

- Warren
 
oh damn... i subtracted 7.5 by 15 instead of adding them.. silly me...

Have an exam tomorrow so need help with this question too:

Highway safety engineers build soft barriers so that cars hitting them will slow down at a safe rate. A person wearing a seat belt can withstand an acceleration of -3.0 x 10^2 m/s^2. How thick should barriers be to safely stop a carthat hits a barrier at 110km/h?

dunno which equation to use... i know i have to fiind D but i don't know the Vinitial!
 
The initial velocity is 110 km/hr, a given.

How will the car's velocity change when it's undergoing an acceleration of -300 m/s^2?

- Warren
 
im lost... how do i find that out?
 
What does it mean when we say an acceleration is, for example, "300 meters per second per second?"

It means that for each second an object is subjected to that acceleration, it will experience a change in velocity of 300 meters per second.

If your car starts out going 110 km/hr, which is just about 30.5 m/s, and is subjected to an acceleration of -300 m/s^2, then it will take just about a tenth of a second for the car to come to a stop. (In an entire second, the car would lose 300 m/s of velocity; every tenth of a second, the car will lose one tenth of 300 m/s of velocity.)

How far does the car go in that tenth of a second? Use the standard equation:

[tex]s(t) = v_0 t - \frac{1}{2} a t^2[/tex]

plugging in 30.5 for [itex]v_0[/itex], -300 for a and 0.1 for t.

- Warren
 
You are told the initial velocity is 110 km/hr. You are also told the acceleration is -300 m/s^2. If the car is to come to a complete stop, then what is its final velocity? Now can you find the displacement?
 
chroot said:
What does it mean when we say an acceleration is, for example, "300 meters per second per second?"

It means that for each second an object is subjected to that acceleration, it will experience a change in velocity of 300 meters per second.

If your car starts out going 110 km/hr, which is just about 30.5 m/s, and is subjected to an acceleration of -300 m/s^2, then it will take just about a tenth of a second for the car to come to a stop. (In an entire second, the car would lose 300 m/s of velocity; every tenth of a second, the car will lose one tenth of 300 m/s of velocity.)
- Warren

"in an entire second SQUARED not just a second"... its not just a second right?? its m/S^2... so would the value of 1/10 still remain the same takin into consideration the S being squared?
 
There's no such thing as a second squared. Read this sentence over again:
It means that for each second an object is subjected to that acceleration, it will experience a change in velocity of 300 meters per second.
In one second, the car loses 300 m/s of velocity. That's an acceleration of -300 meters per second, per second.

- Warren
 

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