Physics question help catching bus

In summary: Raj's equation of motion is $$v_o = -d_o / (t-\tau)$$where ##t = 0,\tau = 1.0s##.At time t, Raj's position is $$v_o = 25 + (-d_o/t)$$which is the same as the bus' position.In summary, Raj is trying to catch a bus that is accelerating away from him. If he catches the bus, he will be close to it at the time. However, if he does not catch the bus, he will be far away from it at the time.
  • #1
33639
12
0
Hi! I'm have difficulty with this question. If anyone can help it would be a lot of help.

Problem
Raj is late for the bus, and as he closes the door to his house, he notices the bus, located 25m away, has begun to accelerate away from him at 1.0 m/s2. Raj begins to chase the bus immediately at a speed of 3.4 m/s. Is Raj able to catch the bus? If he does, where does he catch up to it? If he does not, calculate Raj's frustration distance.

What I've tried to solve
dbus = draj
1/2(v)t2 = vt
1/2(1.0)t2 = 3.4t
0=-1/2t2 + 3.4 t
Used quadratic formula to get
t = -0.25 OR t = 7.08

But at 7.08s Raj is no where near the bus because it takes him 7.352 seconds to reach 25m the distance the bus was originally from Raj...
It doesn't make any sense!

Another question
What is frustration distance? and how can it be calculated?
 
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  • #2
Remember that the bus is initially 25m away from Raj. So the bus' position begins at distance ##d_o = 25m## when Raj's position is still at zero.

Can you write the expression for the bus' position with respect to time given that its initial position is 25m?
 
  • #3
gneill said:
Remember that the bus is initially 25m away from Raj. So the bus' position begins at distance ##d_o = 25m## when Raj's position is still at zero.

Can you write the expression for the bus' position with respect to time given that its initial position is 25m?

I'm not entirely sure i understand what you are asking but would this expression be
25 = vt
??...
 
  • #4
33639 said:
I'm not entirely sure i understand what you are asking but would this expression be
25 = vt
??...

No, the bus' velocity multiplied by time is not a constant 25m.

What is the general formula for accelerated motion, incorporating initial position, velocity, and acceleration?
 
  • #5
gneill said:
No, the bus' velocity multiplied by time is not a constant 25m.

What is the general formula for accelerated motion, incorporating initial position, velocity, and acceleration?

Hmm right it wouldn't be constant 25 because its accelerating.

this formula has d, v, t, a
d= vt + 1/2at2
 
  • #6
33639 said:
Hmm right it wouldn't be constant 25 because its accelerating.

this formula has d, v, t, a
d= vt + 1/2at2

Not quite; you've left out the initial distance ##d_o##. Also, you should recognize that the 'v' in the equation is the initial velocity, ##v_o##.

So, the distance of the bus from Raj's door is given by

$$d_{bus} = d_o + v_o t + \frac{1}{2}a t^2 $$
where ##d_o = 25m##, ##v_o = 0 m/s## and ##a = 1.0 m/s^2##.

Thus you have the equation of motion for the bus with respect to Raj's door. Write Raj's equation of motion similarly, then see if they will 'meet' at some time t.
 

1. How does the physics of motion play a role in catching a bus?

The physics of motion is crucial in catching a bus. It involves understanding concepts such as acceleration, velocity, and distance. When catching a bus, one must move with enough speed and in the right direction to reach the bus before it leaves the stop. This requires applying the principles of motion to calculate the necessary speed and direction.

2. How does the weight of a person affect their ability to catch a bus?

The weight of a person does not significantly affect their ability to catch a bus. However, it does play a role in determining the amount of force needed to accelerate and move towards the bus. A heavier person may require more force to reach the bus in time compared to a lighter person.

3. What is the difference between speed and velocity when catching a bus?

Speed and velocity are often used interchangeably, but they have different meanings in physics. Speed is a measure of how fast an object is moving, while velocity is the speed of an object in a specific direction. When catching a bus, one must consider both speed and velocity to ensure they reach the bus on time.

4. How does the distance between the bus stop and the person affect their ability to catch the bus?

The distance between the bus stop and the person is a crucial factor in catching a bus. The farther away the person is from the bus stop, the longer it will take them to reach the bus. This means they will need to move with a higher speed to cover the distance in time. Understanding the distance is important in calculating the necessary speed and direction to catch the bus.

5. How can Newton's laws of motion be applied when trying to catch a bus?

Newton's laws of motion can be applied when catching a bus. The first law states that an object will remain at rest or in motion with a constant velocity unless acted upon by an external force. In this case, the external force is the person's movement to catch the bus. The second law states that the acceleration of an object is directly proportional to the force applied to it and inversely proportional to its mass. This means that a person needs to apply enough force to accelerate towards the bus and reach it in time. The third law states that for every action, there is an equal and opposite reaction. This means that the force applied by the person to reach the bus also results in the bus exerting an equal and opposite force on the person, helping them reach the bus faster.

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