You make a trip in three phases. You first drive for 2.00 hours with an average

Question

You make a trip in three phases. You first drive for 2.00 hours with an average speed of 60.0 mph. You then drive for another 2.00 hours with an average speed of 75.0 mph. For the third phase you drive an additional 200 miles. What should be your average speed on this third phase of your trip if you average speed for the entire trip is to be 70.0 mph? A spring gun is shot horizontally from a desk top. The projectile starts off at a distance H above the floor. The range of the projectile is observed to be 4H. (That is, the projectile moves a distance along the x-axis of 4H before hitting the floor.) What angle measured from the positive x-axis will result in the a range of 6H, if the gun is again shot from the same desk top with the same initial velocity and the projectile again starting off at a distance H above the floor?

Explanation / Answer

Distance travel in first two hour = speed*time = 60*2 = 120 miles

Distance travel in next two hour = speed*time = 75*2 = 150 miles

Distance travel in third phase = 200 miles

total distance travel = 470 miles

average speed of entire journey = Distance / time

70 = 470/time

time = 6.714 hour

that means the last 200 miles is travel in 2.714 hour { 6.714-4 = 2.714 hour}

so average speed to travel 200 miles = Distance/ time = 200 / 2.714 = 73.684 mph

2.)net vertiacl displacement = -H

using equn of motion

S = ut + 0.5 at^2

-H = vsin(theta)*t - 0.5*9.81 t^2 ...........1

in horizontal direction

S = ut + 0.5at^2

6H = vcos(theta) *t

t = 6H/vcos(theta) .......2

put the value of equn 2 into equn 1

-H = vsin(theta)*[6H/vcos(theta)] - {0.5*9.81 [6H/vcos(theat)]^2}

-1 = 4* tan(theta) - 176.58H/(vcos(theta)^2

1- sin(theta)^2 + 2sin(2theta) = 176.58H/v

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