How to Calculate the Distance of a Trip with a Rest Stop and Constant Speed?

  • Thread starter Thread starter Embison
  • Start date Start date
AI Thread Summary
To calculate the distance of a trip with a rest stop and constant speed, the person drove at 89.7 km/h but took a 21.5-minute break, resulting in an average speed of 70.9 km/h. The equations derived show that the total time of the trip was 1.71 hours, leading to a distance of 121 km. The calculations involved setting up equations for distance based on speed and time, factoring in the rest period. The discussion highlights the importance of understanding basic algebraic manipulation to solve such problems. The final conclusion confirms that the trip was 121 km long.
Embison
Messages
49
Reaction score
0
The question is:

A person takes a trip, driving with a constant speed 89.7 km/h except for a 21.5 min rest stop.
If the person's average speed is 70.9 km/h, how far is the trip? Answer in units of km.



I haven't done this kind of thing for years! and I was just wondering if I solved it properly.. here is what i came up with...



time = 89.7 x (t - 21.5/60) = 70.9, 89.7 t - 38.5 = 70.9 t, t = 2.70 h

distance = 70.9 x 2.70 = 191.43 km


Is that right? Any help would be greatly appreciated.

Thanks
 
Mathematics news on Phys.org
Let d be the length of the trip (km). Let t be the time of the trip (hours).

<br /> 70.9t = d<br />
<br /> 89.7(t - 21.5/60) = d<br />
<br /> 70.9t = 89.7t - 1928.55/60<br />
<br /> 18.8t = 1928.55/60<br />
<br /> t = 1.71<br />
<br /> d = 70.9t = 121<br />

The trip was 121 km long and took 1.71 hours.
 
i wish i knew how you did that...

but i don't even know what / <-- means?

i think it means to divide but I've forgotten all of the math i was taught.

it's been years.
 
Okay, here's a "translation".

As Tron3k said "Let d be the length of the trip (km). Let t be the time of the trip (hours)."

Since the person stopped for 21.5 min.= 21.5 min/60 min/h (21.5 divided by 60)= 0.358 h, the time actually in motion (at a constant speed of 89.7 km/h) was t- 0.358 hours. Since speed is defined as "distance divided by time" we have 89.7= d/(t-0.358) or (multiply by t-0.358 to get rid of the fraction) 89.7(t-0.358)= d.

We are told that the average speed (averaging the constant 89.7 when he was moving with the 0 speed while stopped) was 70.9 km/h.
The average speed is calculated (again using "distance divided by time") using the total time: 70.9= d/t or d= 70.9t.

Since d is the same in both equations, d= 70.9t= 89.7(t- 0.358)=
89.7t- (89.7)(0.358)= 89.7t- 32.15 or
70.9t= 89.7t- 32.15

Subtract 89.7t from both sides: -18.8t= -32.15. Now divide both sides of the equation by -18.8: t= -32.15/-18.8= 1.71 hours.
(Much to my amazement, the same answer Tron3k got!)

Now put that back into either d= 89.7(t- 0.358) or d= 70.9t:
Since d= 70.9t is easier: d= 70.9(1.71)= 121 km.
 
Originally posted by Embison
i wish i knew how you did that...

but i don't even know what / <-- means?

i think it means to divide but I've forgotten all of the math i was taught.

it's been years.



hahaha rofl :wink:
 
I wanted to ask how you got 18t?
 
K_go_1 said:
I wanted to ask how you got 18t?
? What "18t"? I did get "18.8t" exactly like I said:

70.9t= 89.7t- 32.15

Subtract 89.7t from both sides
70.9t- 89.7t= (70.9- 89.7)t= -(89.7- 70.9)t= -18.8t.
That made the equation -18.8t= -32.15 so I multiplied both sides by -1:

18.8t= 32.15.

How did you find this thread? It is 7 years old!
 
Back
Top