Speed, Velocity and Acceleration Problems with Two Objects Involved

[CallMeMarvin]

Homework Statement

1. A person takes a trip, driving with a constant speed of 89.5 km/h, except for a 22 min rest stop. If the person's average speed is 77.8 km/h, how much time is spent on the trip and how far does the person travel?

2. Two sprinters finish with times of 3:53.58 and 3:55.66. Assuming that both run 1609 m at constant velocity,, what distance separates them at the end of the race?

Homework Equations

S = d/t?

I don't know but I sort of feel they're the ones involve here. PLease help me, cause I don't have any idea with this, and its our exam tomorrow

The Attempt at a Solution

1. I did a tabular there.
1st Part has only speed given.
2nd Part has only the time. [rest for a moment]
3rd Part has the total speed.

but got stuck when solving it, cause I can't find the distance and time of the 1st velocity. I tried diving the time of the 2nd to the speed of the first but it doesn't seem to work out.

2. For here so far
i just converted the two times into seconds and then I each multiplied them to their designated speeds but got stuck in what i really need to find.

Sorry :( i really need your help, can you please please please help me :(((
i'm sorry if this is all i can give for an attempt. :'(

 

[gneill]

Homework Statement

1. A person takes a trip, driving with a constant speed of 89.5 km/h, except for a 22 min rest stop. If the person's average speed is 77.8 km/h, how much time is spent on the trip and how far does the person travel?

2. Two sprinters finish with times of 3:53.58 and 3:55.66. Assuming that both run 1609 m at constant velocity,, what distance separates them at the end of the race?

Homework Equations

S = d/t?

I don't know but I sort of feel they're the ones involve here. PLease help me, cause I don't have any idea with this, and its our exam tomorrow

The Attempt at a Solution

1. I did a tabular there.
1st Part has only speed given.
2nd Part has only the time. [rest for a moment]
3rd Part has the total speed.

but got stuck when solving it, cause I can't find the distance and time of the 1st velocity. I tried diving the time of the 2nd to the speed of the first but it doesn't seem to work out.

Assign a variable name, say T, to the unknown total time for the trip. Can you write an expression for the total distance traveled if the driving speed is V and the rest stop time Tr?
(Hint: If the total time is T and the rest time is Tr, how much time was spent at speed V?)

2. For here so far
i just converted the two times into seconds and then I each multiplied them to their designated speeds but got stuck in what i really need to find.

Given the track length and the finish time of the second runner, what was his speed?
How much time separated the second runner's finish from the first runner's finish?

 

[CallMeMarvin]

1st Question..

so this should be like S = VT_r ?
2nd Question..

2nd Runner's speed is 6.85m/s
1st Runner's speed is 6.91m/s

2nd Runner's time is 235 seconds
1st Runner's time is 233 seconds

so tell me if i follow.
then i should find the difference of the two times so that's 2 seconds and then what?
should i multiply 2 seconds to the 2nd runner speed then if so, I obtained 13.7m. and same goes with the 1st runner, which I have obtained a distance of 13.8m.

Should I then subtract them? so the actual distance of the 1st to the Runner is .1m ?

 

[gneill]

1st Question..

so this should be like S = VT_r ?

Well, not quite. He's not moving at all while he's resting, so he travels no distance during time Tr. He DOES move during the rest of the time... So if T is the total time and Tr is the rest time, what is the time during which he is moving?

2nd Question..

2nd Runner's speed is 6.85m/s
1st Runner's speed is 6.91m/s

2nd Runner's time is 235 seconds
1st Runner's time is 233 seconds

so tell me if i follow.
then i should find the difference of the two times so that's 2 seconds and then what?
should i multiply 2 seconds to the 2nd runner speed then if so, I obtained 13.7m.

Stop right there. At the instant the first runner finishes the second runner still has 2 seconds to run according to your calculations. He does so at the speed you calculated for him. So the distance you calculated IS the distance between the runners.

 

[CallMeMarvin]

for the 1st question.

i actually don't get what you mean there :( .. sorry.
is it

T_m [m for moving] = T - T_r
where in the V's time is T_m?

cause as of now, we cannot obtain the total time and the V's time right?

for the 2nd question.

Sorry. but am i right? when i calculated the speed of the 2nd runner then multiplied their difference of time [which is 2 seconds] with it? i'd actually get the distance of the two? :D

presenting the mathematical representation:

S_{2nd Runner} = d_{of runners} / t_{where it is their difference between the time they ran the whole lap}

 

[gneill]

for the 1st question.

i actually don't get what you mean there :( .. sorry.
is it

T_m [m for moving] = T - T_r
where in the V's time is T_m?

cause as of now, we cannot obtain the total time and the V's time right?

As of now you can write the total time that the person is in motion as: T - T_r, where T is still unknown. Write the total distance he drives given that his speed is V=89.5 km/h over the time (T - Tr). Then consider what the definition of average speed is (and since you're given a value for the average speed you should then be in a position to solve for the total time, T).

for the 2nd question.

Sorry. but am i right? when i calculated the speed of the 2nd runner then multiplied their difference of time [which is 2 seconds] with it? i'd actually get the distance of the two? :D

presenting the mathematical representation:

S_{2nd Runner} = d_{of runners} / t_{where it is their difference between the time they ran the whole lap}

When the first runner is at the finish line the second runner still has approximately 2 seconds of running to do at his constant speed. You can work out the distance from the time and speed. NOTE: be sure to calculate the time difference accurately. Hundredths of a second are important here.