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Readme [11.4K]
3 years ago
9

An automobile travels on a straight road for 42 km at 45 km/h. it then continues in the same direction for another 42 km at 90 k

m/h. (assume that it moves in the positive x direction.) a. what is the average velocity of the car during this 84 km trip?
b. what is it’s average speed?
C. Graph x versus t and indicate how the average velocity is found on the graph?

Physics
1 answer:
Nina [5.8K]3 years ago
5 0

Answer:

a) The average velocity is v = (60 km/h ; 0)

b) The average speed is 60 km/h

Explanation:

The velocity is a vector that has a magnitude and direction. The average speed is the distance traveled over time without taking into account the direction of the motion.

a)The average velocity is calculated as the displacement over time:

v = Δx/Δt

where

v = velocity

Δx = final position - initial position = traveled distance relative to the center of the reference system.

Δt = final time - initial time (initial time is usually = 0)

We know that the displacement is 84 km but we do not know the time. It can be calculated from the two parts of the trip.

In part 1:

v = 45 km/h = 42 km / t

t = 0.93 h

In part 2:

v = 90 km/h = 42 km / t

t = 42 km / 90 km/h

t = 0.47 h

The time of travel is 0.47 h + 0.93 h = 1.4 h

The average velocity will be:

v = 84 km / 1.4 h = 60 km/h

Expressed as a vector in a 2-dimension plane:

v = (60 km/h; 0)

b) The average speed is calculated as the distance traveled over time. Note that in this case, the distance is equal to the displacement since the direction of the motion is always in one direction. But if the direction of the second part of the trip would have been the opposite to the direction of the first part, the displacement would have been 0 (final position - initial position = 0, because final position = initial position), then, the average velocity would have been 0. In change, the average speed would have been the distance traveled (84 km, 42 km in one direction and 42 km in the other) over time.

Then:

average speed = 84 km / 1.4 h = 60 km/h

c) see attached figure.    

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a(t) = -9.8m/s^2

To get the vertical velocity we need to integrate over time to get:

v(t) = (-9.8m/s^2)*t + v0

where v0 is the initial vertical velocity because the object is thrown horizontally, we do not have any initial vertical velocity, then v0 = 0m/s

v(t) = (-9.8m/s^2)*t

To get the vertical position equation we need to integrate over time again, to get:

p(t) = (1/2)*(-9.8m/s^2)*t^2 + p0

where p0 is the initial position, in this case is the height of the edifice, 20m

then:

p(t) = (-4.9m/s^2)*t^2+ 20m

The projectile will hit the ground when p(t) = 0m, then we need to solve:

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The correct option is a.

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Now we can use the relationship:

distance = speed*time

We know that the projectile travels for 2.02 seconds, then the total distance that it travels is:

distance = 2.02s*25m/s = 50.5m

Here the correct option is a.

c) Again, the horizontal velocity never changes, is 25m/s constantly, then here the correct option is option b. 25m/s

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The velocity is negative because it goes down, and it matches with option d, so I suppose that the correct option here is option d (because the sign depends on how you think the problem)

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