Solving a Physics Word Problem: Bob & Jack's 1000km Race

[aisha]

Bob and Jack enter a 1000 km bike race. Bob cycles 4 km/h faster than Jack, but his bike gets a flat tire, which takes 30 minutes to repair. If the both of them finish the race in a tie, then how fast was each boy cycling?

Wooooh I have no clue as to how to solve this problem, can someone help me Please!

 

[learningphysics]

let x=Jack's speed then, Bob's speed=x+4. Write an equation for the time Jack takes to finish the race in terms of x. Then write an equation for the time Bob takes to finish the race in terms of x.

Since the two times are equal, you can set them equal and solve for x.

 

[aisha]

ahhh I am really confused I don't understand what the equations are the 30 min to repair is throwing me off. Jack is 4km less than bob +30 minutes? I am still VERY STUCK HELP

 

[bobp718]

The previous post had the right idea. What you need to say is that if they completed the race at the same time then their time in transit is the same, in other words [tex]T_{Bob}=T_{Jack}[/tex] however their time in motion is not the same. a simple system of two equations will satisfy your problem. Both describe the motion of the the racers.

[tex]V_{Bob}T=1000 \rightarrow (V_{Jack}+4)(T-.5)=1000[/tex]
[tex]V_{Jack}T=1000 \rightarrow T=\frac{1000}{V_{Jack}[/tex]

at this point you can make a simple substitution of the bottom equation into the other to get

[tex](V_{Jack}+4)(\frac{1000}{V_{Jack}-.5)=1000[/tex]

this reduces to the quadratic

[tex]-.5V_{Jack}^2-2V_{Jack}+4000=0[/tex]

this should give you that

[tex]V_{Jack}=87.5[/tex]
[tex]V_{Bob}=91.5[/tex]
[tex]T=11.4[/tex]

If there is funny text in this post, i apologize. Residual text from previous posts are coming back to haunt me for some reason.

 

[learningphysics]

Remember that distance=speed * time ie: time=distance/speed
We have to be careful about units. Since speed is in km/h, I'll convert 30 minutes into hours (0.5 hours)

Think about Bob's time like this:
Bob's total time= Bob's cycling time + Bob's repair time = (1000/(x+4)) + 0.5



Jack's time is: 1000/x
Bob's time is: (1000/(x+4)) + 0.5

The two times are the same so:

[tex]\frac{1000}{x}=\frac{1000}{x+4} + 0.5[/tex]

Solve for x. Hope this helps.

 

[aisha]

Remember that distance=speed * time ie: time=distance/speed
We have to be careful about units. Since speed is in km/h, I'll convert 30 minutes into hours (0.5 hours)

Think about Bob's time like this:
Bob's total time= Bob's cycling time + Bob's repair time = (1000/(x+4)) + 0.5



Jack's time is: 1000/x
Bob's time is: (1000/(x+4)) + 0.5

The two times are the same so:

[tex]\frac{1000}{x}=\frac{1000}{x+4} + 0.5[/tex]

Solve for x. Hope this helps.

Yes That helped soooo much well my LCD was x(x+4) when I expanded and simplified I got a quadratic equation ->0.5x^(2)+2x-4000 I used the quadratic formula and got x=87.5, or x=-91.5 the negative solution is impossible because Jack's time cannot be negative? so if x=jack then (87.5+4) = bob's time of 91.5 km/h therefore Jack was cycling 87.5km/h and Bob was cycling 91.5km/h? Is this correct?

 

[learningphysics]

Yes That helped soooo much well my LCD was x(x+4) when I expanded and simplified I got a quadratic equation ->0.5x^(2)+2x-4000 I used the quadratic formula and got x=87.5, or x=-91.5 the negative solution is impossible because Jack's time cannot be negative? so if x=jack then (87.5+4) = bob's time of 91.5 km/h therefore Jack was cycling 87.5km/h and Bob was cycling 91.5km/h? Is this correct?

Yes. I got the same thing, and so did bobp718

 

[aisha]

Yes. I got the same thing, and so did bobp718

THANKS AGAIN U GUYS WOHOOO I FEEL GREAT AND U SHOULD TOO FOR HELPING SOMEONE THANKS