How Do You Calculate Trip Duration and Distance with Stops and Average Speed?

  • Thread starter Thread starter aquapod17
  • Start date Start date
  • Tags Tags
    Speed Time
AI Thread Summary
To calculate trip duration and distance with stops, the average speed formula v = d/t is essential. The person drives at a constant speed of 90.5 km/h but takes a 22-minute rest, affecting the overall average speed to 73.4 km/h. The equation 90*t = 73.4*(t + 22/60) can be used to relate the distance traveled before and after the stop. By solving this equation, the total time spent on the trip and the distance traveled can be determined. This approach effectively incorporates both driving speed and rest duration into the calculations.
aquapod17
Messages
11
Reaction score
0

Homework Statement


A person takes a trip, driving with a constant speed of 90.5 km/h except for a 22.0 min rest stop. If the person's average speed is 73.4 km/h, how much time is spent on the trip and how far does the person travel?


Homework Equations


v = d/t


The Attempt at a Solution


I have no idea how to solve this, other than that I have to plug the values into the average speed formula.
 
Physics news on Phys.org
aquapod17 said:

Homework Statement


A person takes a trip, driving with a constant speed of 90.5 km/h except for a 22.0 min rest stop. If the person's average speed is 73.4 km/h, how much time is spent on the trip and how far does the person travel?

Homework Equations


v = d/t

The Attempt at a Solution


I have no idea how to solve this, other than that I have to plug the values into the average speed formula.

From your equation you know that d = V*t

If you hadn't stopped you know that 90.5*t is the distance
But you did stop 22 minutes (which is 22/60 hours) so to do the same distance it was 73.4*(t+22/60)

Since the distance is the same then 90*t =73.4*(t+22/60)

I'm sure you get the idea from here.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top