Física

Chapter 2 – Student Solutions Manual
1. We use Eq. 2-2 and Eq. 2-3. During a time tc when the velocity remains a positive constant, speed is equivalent to velocity, and distance is equivalent to displacement, with
Δx = v tc.
(a) During the first part of the motion, the displacement is Δx1 = 40 km and the time interval is t1 =

(40 km)
= 133 h.
.
(30 km / h)

During the second part the displacement is Δx2 = 40 km and the time interval is t2 =

(40 km)
= 0.67 h.
(60 km / h)

Both displacements are in the same direction, so the total displacement is
Δx = Δx1 + Δx2 = 40 km + 40 km = 80 km.
The total time for the trip is t = t1 + t2 = 2.00 h. Consequently, the average velocity is vavg =

(80 km)
= 40 km / h.
(2.0 h)

(b) In this example, the numerical result for the average speed is the same as the average velocity 40 km/h.
(c) As shown below, the graph consists of two contiguous line segments, the first having a slope of 30 km/h and connecting the origin to (t1, x1) = (1.33 h, 40 km) and the second having a slope of 60 km/h and connecting (t1, x1) to (t, x) = (2.00 h, 80 km). From the graphical point of view, the slope of the dashed line drawn from the origin to (t, x) represents the average velocity.

5. Using x = 3t – 4t2 + t3 with SI units understood is efficient (and is the approach we will use), but if we wished to make the units explicit we would write x = (3 m/s)t – (4 m/s2)t2 + (1 m/s3)t3.
We will quote our answers to one or two significant figures, and not try to follow the significant figure rules rigorously.
(a) Plugging in t = 1 s yields x = 3 – 4 + 1 = 0.
(b) With t = 2 s we get x = 3(2) – 4(2)2+(2)3 = –2 m.
(c) With t = 3 s we have x = 0 m.
(d) Plugging in t = 4 s gives x = 12 m.
For later reference, we also note that the position at t = 0 is x = 0.
(e) The position at t = 0 is subtracted from the position at t = 4 s to find the displacement
Δx = 12 m.
(f) The position at t = 2 s is subtracted from the