How Do You Calculate Car Speed When Dodging a Water Balloon?

[jim_ringo]

Homework Statement

You are driving at a constant speed on a straight highway that passes under an overpass. At time , when your car is a distance from the overpass, a prankster on the overpass releases a water balloon from rest from a height above the highway below.

Answer the following questions. You should treat the car as a point particle, ignoring its height, length, and width.

(a) Suppose that you don’t see the water balloon until it explodes on your windshield. At what speed were you originally traveling, given that the water balloon strikes your car? Express your answer in terms of d, h, and any relevant constants. Simplify your answer as much as possible.

(b) Now suppose that you see the prankster release the water balloon at t=0 , and you slam on your brakes at time in an attempt to avoid being hit. (Your reaction time is Tb>0 .) Your car slows down with a constant acceleration of magnitude . The balloon hits the highway right in front of your car at the instant your car comes to rest. At what speed Vo were you originally traveling, given that the water balloon just barely misses your car? Express your answer in terms of d,h As,Tb and any relevant constants. Simplify your answer as much as possible.

(c) Let Va be the speed you found in part (a), and let Vb be the speed you found in part (b). Which of these two speeds is larger? Justify your answer mathematically, with comments as necessary.

Homework Equations

x=vt
v=at
x=at^2

The Attempt at a Solution

Acceleration of balloon= 9.8m/s^2 or -g.

So balloon time to hit ground is t=sqrt(h/-g)

d= distance car travels to hit the balloon

d=Vo sqrt(h/-g) therefore Vo = d/(sqrt(h/-g))

For part b I am not sure how to account for the reaction time.

Thanks

 

[haruspex]

x=at^2

That's not quite right, and has led you to a wrong answer.

For part b I am not sure how to account for the reaction time.

Just develop some equations. Let the speed be v₀. How far will you travel during your reaction time? How much further while braking?

 

[jim_ringo]

corrections

I have made some corrections to my work and would like to know if I am correct.

I used the formula x-x naught= Vnaught t +1/2at^2

Balloons accelration= g
Height= h
Velocity of car=Vnaught

The Vnaught of the balloon is zero.

So I said

h=Vnaugh t +1/2 gt^2 therefore t = sqrt(2h/g)= Time the balloon takes to fall.

The car must travel (d) on the time the balloon takes to fall (t).
The time the car traveled would be d/Vnaught
So I plugged d/Vnaught into h=1/2 gt^2

This led me to calculate that Vnaught= d sqrt(g/2h)

Is this correct?


For part B I let t=sqrt(2h/g)= time the balloon takes to fall

As= deceleration of the car
Tb= the reaction time.

The distance the car moves is the overall distance minus the velocity initial times the braking time.

so d=(d-Vnaught Tb)

The final velocity of the car (V) is zero.

so I said d=Vnaught sqrt(2h/g)+1/2 (As) 2h/g...correct or not?

 

[haruspex]

I used the formula x-x naught= Vnaught t +1/2at^2

Good.

Balloons accelration= g
Height= h
Velocity of car=Vnaught

The Vnaught of the balloon is zero.

So I said

h=Vnaugh t +1/2 gt^2 therefore t = sqrt(2h/g)= Time the balloon takes to fall.

Please avoid using labels generically. The speeds of the car and balloon are different, so give them different names.

The car must travel (d) on the time the balloon takes to fall (t).
The time the car traveled would be d/Vnaught
So I plugged d/Vnaught into h=1/2 gt^2

This led me to calculate that Vnaught= d sqrt(g/2h)

Yes

For part B I let t=sqrt(2h/g)= time the balloon takes to fall

As= deceleration of the car
Tb= the reaction time.

The distance the car moves is the overall distance minus the velocity initial times the braking time.

so d=(d-Vnaught Tb)

Again, label the two distances differently.

The final velocity of the car (V) is zero.

so I said d=Vnaught sqrt(2h/g)+1/2 (As) 2h/g...correct or not?

I can't be certain because I don't know which version of d this is, but it could be.

 

[jim_ringo]

Good.
Please avoid using labels generically. The speeds of the car and balloon are different, so give them different names.
YesAgain, label the two distances differently.
I can't be certain because I don't know which version of d this is, but it could be.

That version of d is the distance the car moves minus the initial velocity times the braking time d=(d-Vnaught Tb) Therefore am I correct?

Thanks

 

[haruspex]

That version of d is the distance the car moves minus the initial velocity times the braking time d=(d-Vnaught Tb) Therefore am I correct?

Thanks

Yes (but you mean reaction time; braking time is the time during braking).

 

[akh]

Why is the time used in the last equation t rather than (t-tb)?

 

[haruspex]

Why is the time used in the last equation t rather than (t-tb)?

Good catch, thanks.

 

[jim_ringo]

c) Let Va be the speed you found in part (a), and let Vb be the speed you found in part (b). Which of these two speeds is larger? Justify your answer mathematically, with comments as necessary.

For Va I have Va=dsqrt(g/2h)

For Vb I have Vb=(dg-Ash)/(g(2sqrt(2h/g)-tsub b))

How can I determine which is larger? Can a kinematic equation by applied?

Thanks

 

[haruspex]

For Vb I have Vb=(dg-Ash)/(g(2sqrt(2h/g)-tsub b))

That doesn't look right. Please show your working to that point, preferably using clearer notation. Try using subscripts, sqrt symbol, etc.