Physics question help catching bus

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  • #1
33639
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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?
 

Answers and Replies

  • #2
gneill
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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
33639
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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
gneill
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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
33639
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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
gneill
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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.
 

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