Question

When to add and subtract with velocity.

Answer 1

I solved these two question but in the first one i added my time and position
in the second one i had to subtract my time and position. Why???



1.)(a) You walk 75.7 m at a speed of 1.22 m/s and then run 75.7 m at a speed of 3.05 m/s along a straight track.

2.)The position of a particle moving along the x axis is given in centimeters by x = 8.25 + 3.50t 3, where t is in seconds. Consider the time interval t = 2.00 s to t = 3.00 s.
(a) Calculate the average velocity.

Answer 2

1.)
d=r*t
75.7m = 1.22 m/s *t 75.7m=3.05m/s *t
t^1=62.049s t^2=24.82s
75.7+75.7= 151.4m
62.049+24.82s=86.874s

velocity= 151.4/86.874 = 1.74 m/s

Answer 3

x(t)=8.25+3.50t^3
plugged in 3 = 102.75
plugged in 2 = 36.25
t^1=2
t^2=3

(102.75-36.25)/(3-2)= 66.5 cm/s

Answer 4

why was i allowed to add in the first one but i had to subtract in the second one.
thanks for the help in advance .

Answer 5

Well, let's see.
In the first one you're told you've walked \(75.7_m\) at one given speed \(1.22_{m/s}\) and then ran another \(75.7_m\) at another speed \(3.05_{m/s}\).
That means (\(\text{total_trip_time} = \text{time_of_walk} + \text{time_of_run}\))
and (\(\text{total_trip_distance} = \text{distance_of_walk} + \text{distance_of_run}\))
after calculating those 2 you could calculate the average trip velocity by
$$
\text{average_trip_velocity} = \frac{\text{total_trip_distance}}{\text{total_trip_time}}
$$
In the second one however, you have an equation relating the time passed to the position of the particle along the 'x' axis.
Here what you are told to find out how long the particle has traveled between t=2 and t=3.
Here again we need to calculate the average trip velocity by:
$$
\text{average_trip_velocity} = \frac{\text{total_trip_distance}}{\text{total_trip_time}}
$$
We have to find the same things as with the previous question, but we get different information.
In this question (unlike the previous one) we are told the time at the beginning of the trip and the time at the end. By subtracting those you get the time in between, which is the time of the trip.

Similarly we are only know the 'position' of the particle at the beginning and at the end of the trip and we have to extract the distance it has passed.
Luckily the particle is moving in a single direction so by knowing the position at the end of the trip we know how much it has moved since the beginning of the trip.
If the particle could go backwards then we would have a problem because the change in its position wouldn't reflect the distance it has traveled. (it could have returned to the same position it started, but distance is not 0 for example)

Hope it helps =S