Section 2.2  Speed and Velocity

1. (Exercises 5 and 6) A student travels from school to her home in 8.5 minutes. If the total distance that she traveled is 1.4 km, what is her average speed in m/s?

First we must convert the given distance and time values to the proper SI units. 8.5 min (60 s / 1 min) = 510 s 1.4 km (1000 m / 1 km) = 1400 m Now we can apply the relation defining average speed to this data. speed = distance / time = 1400 m / 510 s = 2.745 m/s = 2.7 m/s

2. (Exercises 7 and 8) A bird flies to a point that is 4100 km due south of its starting point in 5.0 days, what is the average velocity for this trip.

There are no specific units asked for, so we will find the velocity in km/h. The displacement is already given in km, but we must convert days into h. 5.0 days (24 h / 1 day) = 120 h The average velocity is then: v = d / t = 4100 km / 120 h = 34.16 km/h = 34 km/h due south

Notice that we have rounded the answer to 2 sig. fig., as required, and included the direction. If we had chosen m/s as the derived units, the average velocity would have been 9.5 m/s. Try this and see if you can do the conversion.

Section 2.3  Acceleration

3. (Exercises 9 and 10) How long does it take for a speedboat to decelerate from 35.0 m/s to 9.0 m/s if the average deceleration of the boat is -12 m/s²?

Starting with the defining equation for acceleration, we solve for time and obtain:

a = v/t sot = (v_F - v_o ) / a

This means that the deceleration of the speedboat took place in:

t = (9.0 m/s - 35.0 m/s) / -12 m/s² = 2.2 s

4. (Exercises 15 and 16) A bolt drops from the roof of a new construction project and hits the ground 3.52 s later.

a. How far did the bolt fall before it reached the ground?

b. What was the final velocity of the bolt just before it struck the ground? Because the bolt was in free-fall, it had an acceleration that was equal to -9.80 m/s2, and because it fell from the roof we will assume an initial downward velocity of zero.

a. The displacement of its fall will then be:

d = ¹/₂ a t² = ¹/₂ (-9.80 m/s²) (3.52 s)² = -60.7 m, downward

b. We can use 3 sig.fig. if we use g = -9.80 m/s², since the time was given to 3 sig.fig.

v_F = v_o + a t = 0 + -9.80 m/s² (3.52 s) = -34.5 m/s, downward

Section 2.4  Acceleration in Uniform Circular Motion

5. (Exercises 17 and 18) A rock is swung around in a circle at the end of a piece of string 2.7 m long. If the rock is traveling at a constant speed of 4.0 m/s, what is the acceleration of the rock? Because the rock is traveling with a constant speed and is not slowing down or speeding up, it will not have a linear acceleration, but because it is traveling on a curved path, it will have a centripetal acceleration.

a_c = v ² / r = (4.0 m/s)² / 2.7 m = 5.9 m/s²

Section 2.5  Projectile Motion

6. (Exercises 19 and 20) A wrench is kicked off the edge of a beam that is located 8.5 m above the ground with a horizontal initial velocity of 2.4 m/s. Assume that there is no initial vertical velocity and that friction drag on the falling wrench is negligible. Remember that the vertical acceleration will be g. What will be the vertical (Y) and horizontal (X) components of the wrench's velocity after it has been falling for 2.0 seconds?

First, let's list what we know.

vXo = 2.4 m/s		aX = 0		height = 8.5 m
vYo = 0		aY = g = -9.80 m/s2		t = 2.0 s

Now we can calculate the separate velocities using the common time factor of 2.0 seconds.

v_YF = v_Yo + a_Yt = 0 + (-9.80 m/s²) (2.0 s) = -19.6 m/s = -20 m/s, downward

v_XF = v_Xo + a_Xt = 2.4 m/s + 0 = 2.4 m/s,

in the same direction it was kicked

Note that the downward velocity of the wrench will increase as it falls because of the acceleration due to gravity, but the horizontal velocity will remain the same throughout the fall because there is no acceleration in the horizontal direction. This is the case for all objects in free-fall when air resistance can be neglected.

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