Problem 81.An object travels a distance of5 kmtowards east, then4 kmtowards north and finally10 kmtowards east1) what is total traveled distance?2) what is resultant displacement?Solution:We show each displacement (travelled paths) in the figure.(1) The total traveled distance is the sum of all traveled distances. It does not matter what the relative directions of displacements are, we just need to add the magnitude of displacements:

(2) The second part of the problem is easier to solve in terms of components of vectors.The net (resultant) displacement is the vector sum of three displacements: the first displacement (), the second displacement (), and the third displacement ():

Then we can find the x and y components of the displacement vectors and finally the net displacement:

Then the magnitude of the net displacement is

Problem 82.Can two displacements (vectors) of different magnitudes be combined to give a zero displacement (resultant)?Solution:The sum of two vectors (displacements) is zero only if1. vectors have opposite directions;2. vectors have the same magnitude.Indeed, the sum of two vectors is shown in the figure below for a fixed orientation of the first vector and different orientations of the second vector. In the picture we assume that the magnitude of the second vector is less than the magnitude of the first vector.

We can see that the sum of two vectors has the smallest magnitude when the vectors have opposite directions. In this casethe sum is zero only if the vectors have the same magnitude.Problem 83.What is the speed (m/sec) needed for a stunt driver to launch from a20 degreeramp to land15 maway?What is his maximum height?Solution:Initial speed.This is the projectile motion. There are two sets of equations, which describe the motion of the projectile (stunt driver):Set 1:motion along horizontal axis (axis x see figure). This is the motion with constant velocity. There is only one equation, which describe this motion:..............................(1)Hereand.Set 2:motion along vertical axis (axisy see figure). This is the motion with constant acceleration free fall motion. There are three equations, which describe this motion. Only two equations are independent, but it is convenient to write all three equations:................(2)...........................(3)...........................(4)

We know that the y-coordinate of the final point (point B) is 0 and the x-coordinate of the final point is 15 m. We substitute these values in equations (1) and (2) and obtain

Then from the first equation we can find:

Substitute this expression into the second equation:

From this equation we can easily find the initial velocity:

Maximum height.The condition that the projectile is at the point with the maximum height is that the y-component of its velocity at this point is zero. It is easier to find the maximum height from equation (4). Indeed, we substitutein this equation and obtain:

Then

We know, then we can find the maximum height:

Problem 84.A body moves4 kmtowards East from a fixed pointAand reaches pointB. Then it covers5 kmtowards North and arrives at pointC. Find the distance and directions of the net displacement.Solution:We show two displacements (travelled paths) in the figure.

The net (resultant) displacement is the vector sum of two displacements: the first displacement () and the second displacement ():

The easiest way to find the net displacement is to introduce coordinate system (axes x and y) and then find the x and y components of the net vector-displacement. The x and y components of the displacement () and displacement () are the following:

Then the x and y components of the net displacement is

Then the magnitude of the net displacement is

The direction of the net displacement is characterized by angle(shown in the figure), which can be found from the known x and y components of the net displacement:

Problem 85.A baseball player hits a homerun, and the ball lands in the left field seats, which is120 maway from the point at which the ball was hit. The ball lands with a velocity of20 m/sat an angle of30 degreesbelow horizontal. Ignoring air resistance(A) find the initial velocity and the angle above horizontal with which the ball leaves the bat;(B) find the height of the ball relatively to the ground.Solution:(A) Initial velocity.Without air resistance this is simple projectile motion. In the present problem we do not know initial velocity: we do not know the magnitude of the velocity (speed) and we do not know its direction.There are two sets of equations, which describe the motion of the projectile (ball).Set 1:motion along horizontal axis (axis x see figure). This is the motion with constant velocity. There is only one equation, which describe this motion:................................................(1)Here.Since the motion along the axis x is the motion with constant velocity then the x-component of the velocity is constant. We know the velocity at the final point. Then we can find the x-component of the velocity at the final point:

This x-component of the velocity is equal to the x-component of the initial velocity:...........(2)We also know the x-coordinate of the final point (point B): it is 120 m. We substitute this value in equation (1) and obtain

From this equation we can find the time of travel from point A to point B:

Now we need to analyze the second set of equations.Set 2:motion along vertical axis (axis y see figure). This is the motion with constant acceleration free fall motion. There are three equations, which describe this motion. Only two equations are independent, but it is convenient to write all three equations:.............(3)................................................(4)

Since the initial y-coordinate is zero, then.............................................(5)

We know the y component of the final velocity

This is the y-component of the velocity at the moment of time. We substitute these values in equation (4) and obtain

From this equation we can find the y-component of the initial velocity:

Finally we know the x- and y-components of the initial velocity:

From these expressions we can find the magnitude of the initial velocity and the direction (angle) of the initial velocity:

Now we knowthe initial velocity.(B) Final height.We need to find the final height of the ball (the final y-coordinate). To find the final height we can use equation (3). We just need to substitute the y-component of the initial velocity and the traveled time in this equation:

Problem 86.A body covers1/4journey with a speed of40 km/h,1/2of it with50 km/hand remaining with the speed of60 km/h. Calculate average speed for entire journey.Solution:In this problem we need to use the definition of the average speed. The average speed is equal to the ratio of the total travelled distance and the total traveled time:

We introduce the total traveled distance asand then calculate the total traveled time.We know that 1/4 of the journey a body moves with a speed of 40 km/h. It means that the body moves a distance ofwith the speed of 40 km/h. Then we can find the time of this motion:

Then the body moves 1/2 of the journey with a speed of 50 km/h. It means that the body moves a distance ofwith the speed of 50 km/h. We can find the time of this motion:

Then the body moves the rest of the journey (which is 1/4 of the journey) with a speed of 60 km/h. It means that the body moves a distance ofwith the speed of 60 km/h. We can find the time of this motion:

Then the total traveled time is

Then the average speed is