Application of the equations of motion with constant acceleration

In summary, the problem involves a bicyclist repairing a flat tire and then chasing after a friend who passes by at a constant speed of 4.0 m/s. The bicyclist accelerates at 2.2 m/s^2 until he catches his friend. Using the equations of motion, the time it takes for the bicyclist to catch his friend is 1.9 seconds, he travels a distance of 4.0 meters, and his speed when he catches up is 7.6 m/s. The concept of relative velocity can be applied to make the problem easier.
  • #1
sillybean
18
0
[SOLVED] Application of the equations of motion with constant acceleration

Homework Statement


A bicyclist is finishing his repair of a flat tire when a friend rides by with a constant speed of 4.0 m/s. Two seconds later the bicyclist hops on his bike and accelerates at 2.2 m/s^2 until he catches his friend.

a)How much time does it take until he catches his friend (after his friend passes him)?
b)How far has he traveled in this time?
c)What is his speed when he catches up?


Homework Equations



V=Vo + at

X= Xo + Vot + 1/2at^2

X= Xo + 1/2(Vo + V)t



The Attempt at a Solution


Okay so I tried figuring out the distance that the friend who passed the one doing repairs traveled in 2 seconds. Since his velocity was constant, I reasoned that he traveled 8 meters in 2 seconds. Then I plugged in the acceleration given in the problem along with the distance i figured into the equation: X= Xo + Vot + 1/2at^2. (I made Xo and Vo = zero since the guy doing the repairs starts from rest)

It looked like this: 8.0m = 0+0+ 1/2(2.2m/s^2)t^2
8.0m = 1.1m/s^2(t^2)
7.27s^2=t^2 (took square root of both sides)
t=2.7s

but that was wrong.

I then figured the distance I reasoned was wrong and actually used an equation this time.
I used X= Xo + 1/2(Vo + V)t
X = 0 + 1/2 (4.0 m/s)(2.0s)
X = 4.0 m

I redid my work using this new distance (which I already thought was wrong >:C)
X= Xo + Vot + 1/2at^2
4.0m = 0 + 0 + 1/2(2.2)t^2
4.0 = 1.1t^2
3.6 = t^2 (square root)
t=1.9s

Guess what. It was wrong.

What am I doing wrong?
 
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  • #2
Suppose it has taken a total of t secs from the time when the friend passes him by to the time when he catches up. Then the friend has traveled at a const speed for t secs, whereas the repairing guy has traveled the same dist in (t-2) s. The rest you know.
 
  • #3
sillybean said:

Homework Statement


A bicyclist is finishing his repair of a flat tire when a friend rides by with a constant speed of 4.0 m/s. Two seconds later the bicyclist hops on his bike and accelerates at 2.2 m/s^2 until he catches his friend.

a)How much time does it take until he catches his friend (after his friend passes him)?
b)How far has he traveled in this time?
c)What is his speed when he catches up?


Homework Equations



V=Vo + at

X= Xo + Vot + 1/2at^2

X= Xo + 1/2(Vo + V)t



The Attempt at a Solution


Okay so I tried figuring out the distance that the friend who passed the one doing repairs traveled in 2 seconds. Since his velocity was constant, I reasoned that he traveled 8 meters in 2 seconds. Then I plugged in the acceleration given in the problem along with the distance i figured into the equation: X= Xo + Vot + 1/2at^2. (I made Xo and Vo = zero since the guy doing the repairs starts from rest)

It looked like this: 8.0m = 0+0+ 1/2(2.2m/s^2)t^2
8.0m = 1.1m/s^2(t^2)
7.27s^2=t^2 (took square root of both sides)
t=2.7s

but that was wrong.

I then figured the distance I reasoned was wrong and actually used an equation this time.
I used X= Xo + 1/2(Vo + V)t
X = 0 + 1/2 (4.0 m/s)(2.0s)
X = 4.0 m

I redid my work using this new distance (which I already thought was wrong >:C)
X= Xo + Vot + 1/2at^2
4.0m = 0 + 0 + 1/2(2.2)t^2
4.0 = 1.1t^2
3.6 = t^2 (square root)
t=1.9s

Guess what. It was wrong.

What am I doing wrong?

use concepts of relative velocity...this shall make the problem much easier...
 
  • #4
thanks again you guys are life savers
 

1. How do you calculate the displacement using the equations of motion with constant acceleration?

The displacement can be calculated using the equation: d = ut + 1/2at^2, where d is the displacement, u is the initial velocity, a is the acceleration, and t is the time.

2. What is the difference between velocity and acceleration?

Velocity is the rate of change of displacement over time, while acceleration is the rate of change of velocity over time. In other words, velocity tells us how much an object's position changes, while acceleration tells us how much its velocity changes.

3. How do you determine the final velocity using the equations of motion with constant acceleration?

The final velocity can be calculated using the equation: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

4. Can these equations be used for non-constant acceleration?

No, these equations are specifically for constant acceleration, where the acceleration remains the same throughout the motion. For non-constant acceleration, more complex equations are needed.

5. How do the equations of motion with constant acceleration apply to real-life situations?

These equations are commonly used in physics and engineering to analyze the motion of objects under the influence of a constant force or acceleration. They can be used to understand the motion of objects such as projectiles, vehicles, and roller coasters.

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