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padilas [110]
1 year ago
11

You want to average 90 km/h on a car trip. You cover the first half of the distance at an

Physics
1 answer:
bogdanovich [222]1 year ago
4 0

(a) The average speed must you have for the second half of the trip to meet your goal is 8 km/h.

(b) The value obtained (8 km/h) is not reasonable for the second half of the distance since the first half is 48km/h.

<h3>What is average velocity?</h3>

Average velocity is defined as the change in position or displacement (∆x) divided by the time intervals (∆t) in which the displacement occurs.

average velocity = total distance / total time

v = (d)/(0.5d/v₁ + 0.5d/v₂)

where;

  • v is the average velocity
  • v₁ is the average velocity during the first half
  • v₂ is the average velocity during the second half

90 km/h = (d) / (0.5d/48 + 0.5d/v₂)

90(0.5d/48 + 0.5d/v₂) = d

0.9375d + 0.5d/v₂ = d

d(0.9375 + 0.5/v₂) = d

0.9375 + 0.5/v₂ = 1

0.5/v₂ = 0.0625

v₂ = 0.5/0.0625

v₂ = 8 km/h

Thus, the average speed must you have for the second half of the trip to meet your goal is 8 km/h.

The value obtained (8 km/h) is not reasonable for the second half of the distance since the first half is 48km/h.

Learn more about average velocity here: brainly.com/question/24739297

#SPJ1

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3 years ago
On a drive from one city to? another, victor averaged 39 mph. if he had been able to average 72 ?mph, he would have reached his
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Let the unknown distance be xmiles
x/39-x/72=11hr
72x-39x/2808=11hr
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You stand on a bridge above a river and drop a rock into the water below from a height of 25 m. (Assume no air resistance)
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PART a)

here when stone is dropped there is only gravitational force on it

so its acceleration is only due to gravity

so we will have

a = g = 9.8 m/s^2

Part b)

Now from kinematics equation we will have

y = v_i t + \frac{1}{2} at^2

now we have

y = 25 m

so from above equation

25 = 0 + \frac{1}{2}(9.8 )t^2

t = 2.26 s

Part c)

If we throw the rock horizontally by speed 20 m/s

then in this case there is no change in the vertical velocity

so it will take same time to reach the water surface as it took initially

So t = 2.26 s

Part D)

Initial speed = 20 m/s

angle of projection = 65 degree

now we have

v_x = vcos\theta

v_x  = 20 cos65 = 8.45 m/s

v_y = vsin\theta

v_y = 20 sin65 = 18.13 m/s

PART E)

when stone will reach to maximum height then we know that its final speed in y direction becomes zero

so here we can use kinematics in Y direction

v_f - v_y = at

0 - 18.13 = (-9.8) t

t = 1.85 s

so it will take 1.85 s to reach the top

5 0
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