Mechanics problems involving velocity,acceleration and gravity

In summary: Ok, here's what I got:1: The rear of a bicycle passes a point O on a road waith a velocity of 4 m/s and an acceleration of 2m/s2. Four seconds later the front of a car paases 0 with a velocity of 2m/s and an acceleration of 4m/s2. When and how far from O does the front of the car meet the rear of the bicycle?…1. s1. ut +1/2at2 =4t +1/2(2)t2 =4t + t2The car moves with s = 4t + t
  • #1
Conor11
8
0

Homework Statement


Hello, i have an exam in two days and spent all day trying to solve mechanic questions. Theres 4 in total, I'm having extreme trouble do not have the sufficient time to spend as i have other topics to cover. Would you please be able to help me out. Thanks the questions are as follows.

1: The rear of a bicycle passes a point O on a road waith a velocity of 4 m/s and an acceleration of 2m/s2. Four seconds later the front of a car paases 0 with a velocity of 2m/s and an acceleration of 4m/s2. When and how far from O does the front of the car meet the rear of the bicycle?

2.A stone is dropped from the top of a cliff which is 100m high. At the same time another stone is vertically thrown upwards from the foot of the cliff woth a speed of 40m/s.AT what height and when do the stones meet?

3.A Balloon starts from rest on the ground and rises vertically with a constant acceleration of 2m/s2. After 8 seconds a stone is dropped from the bottom of the balloon. How long does it take the stone to reach the ground.

4. A stone is thrown verticaly upwards from a point with an initial speed of 50m/s. Three seconds later another stone is thrown vertically up from the same pont witha speed of 70m/s. How long after the first stone leaves the ground do the two stones meet? At what height above the ground do the 2 stones meet?



Homework Equations



Linear motion with constant acceleration equations

1. v= u + at
2 s=ut +1/2at(squared)
3. v(squared)=u(squared)+2as



The Attempt at a Solution


1. s1. ut +1/2at(squared)
=4t +1/2(2)t(squared)
=4t + t(squared)

I have no idea. please help
 
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  • #2
"1: The rear of a bicycle passes a point O on a road waith a velocity of 4 m/s and an acceleration of 2m/s2. Four seconds later the front of a car paases 0 with a velocity of 2m/s and an acceleration of 4m/s2. When and how far from O does the front of the car meet the rear of the bicycle?"

You have the formula for the distance given an initial velocity and acceleration rate. When they collide, what can you say about the distances traveled by each?
 
  • #3
they have traveled the same length from O?
 
  • #4
Welcome to PF!

Hi Conor11! Welcome to PF! :smile:

(try using the X2 button just above the Reply box :wink:)
Conor11 said:
1: The rear of a bicycle passes a point O on a road waith a velocity of 4 m/s and an acceleration of 2m/s2. Four seconds later the front of a car paases 0 with a velocity of 2m/s and an acceleration of 4m/s2. When and how far from O does the front of the car meet the rear of the bicycle?

1. s1. ut +1/2at2
=4t +1/2(2)t2
=4t + t2

yup, the bike moves with s = 4t + t2,

and obviously you've also worked out that the car moves with s = 2t + 2t2

but they're not the same t !

so you need to give them different names (eg t1 and t2), and find a third equation relating t1 to t2

what would that be? :smile:

(a lot of these exam problems involve giving things names! :wink:)
 
  • #5
Conor11 said:
they have traveled the same length from O?

Yes, that is correct. But the times differ by 4 seconds so when you equate the distances, one of the times is (t-4). Solve for t, discard the time that makes no sense and use the other to determine the distance. Both equations will give same distance but one is evaluated at (t-4).
 
  • #6
Oh i see , you get two equations, let them equal and then choose the one that makes sense. And then sub it back into the original to get the distance. Cheers Tim and Lawrence. Can you help me with the others?
 
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  • #7
Conor11 said:
2.A stone is dropped from the top of a cliff which is 100m high. At the same time another stone is vertically thrown upwards from the foot of the cliff woth a speed of 40m/s.AT what height and when do the stones meet?

show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
 
  • #8
ok fair enough well:

do you sub in 100,9.8,and 0 in s=ut +1/2at^2 and then you get 100= 4.9t^"

Then, the same with the other stone or what equation do you use?
 
  • #9
*t^2
 
  • #10
Conor11 said:
do you sub in 100,9.8,and 0 in s=ut +1/2at^2 and then you get 100= 4.9t^"

Then, the same with the other stone or what equation do you use?

yes, one equation each stone, using different names for s,

and a third equation relating those two s's …

show us what you get :smile:
 
  • #11
Help with frequency and wave question please.?
Vibrations from a 400hz tuning fork set up a stationaery waves on a string whihc is fixed at both ends. The speed of the waves on the string is 500m/s. The stationery wave has only 5 antinodes along it. What is the length of the string?
 
  • #12
Sorry how do you make a new post on PF?
 
  • #13
Conor11 said:
Sorry how do you make a new post on PF?

Use the "New Topic" button in the upper left of the main Intro Physics forum window.
 

What is the difference between velocity and acceleration?

Velocity is the rate of change of an object's position with respect to time, while acceleration is the rate of change of an object's velocity with respect to time. In simpler terms, velocity measures how fast an object is moving, while acceleration measures how quickly its speed is changing.

How is acceleration affected by gravity?

In mechanics problems, acceleration due to gravity is typically considered a constant value of 9.8 meters per second squared (m/s²) on Earth. This means that an object's acceleration will always be 9.8 m/s² towards the ground, regardless of its mass or size.

Can an object have a constant velocity and non-zero acceleration?

No, an object cannot have a constant velocity and non-zero acceleration at the same time. This is because acceleration is a change in velocity, so if the velocity is constant, there is no change and therefore no acceleration.

How does air resistance affect an object's velocity and acceleration?

Air resistance, also known as drag, is a force that acts in the opposite direction of an object's motion. This means that as an object moves through air, the force of air resistance will decrease its velocity and therefore also affect its acceleration. In some cases, air resistance can even cause an object to have negative acceleration (deceleration).

What is the difference between average velocity and instantaneous velocity?

Average velocity is the total displacement of an object divided by the total time taken, while instantaneous velocity is the velocity of an object at a specific moment in time. In other words, average velocity gives an overall picture of an object's motion, while instantaneous velocity gives a more precise measurement at a specific point in time.

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