ANSWER TO CHAPTER 2 HOMEWORK
1. (a) We find the average speed from
average speed = d/t = (160m + 80m)/(17.0s + 6.8s) = 10.1m/s.
(b) The displacement away from the trainer is 160m - 80m = 80m; thus the average velocity is
vav = Dx/Dt (80m)/(17.0s + 6.8s) = +3.4m/s, away from trainer.
2. (a) We find the position from the dependence of x on t: x =2.0m - (4.6m/s)t + (1.1m/s2)t2.
x1 = 2.0m - (4.6m/s)(1.0s) + (1.1m/s2)(1.0s)2 = -1.5m;
x2 = 2.0m- (4.6m/s)(2.0s) + (1.1m/s2)(2.0s)2 = -2.8m;
x3 = 2.0m - (4.6m/s)(3.0s) + (1.1m/s2)(3.0s)2 = -1.9m.
(b) For the average velocity we have
vav1-3 = Dx/Dt = [(-1.9m - (-1.5m)]/(3.0s - 1.0s) = -0.2m/s (toward -x);
(c) We find the instantaneous velocity by differentiating:
v = dx/dt = -(4.6m/s) + (2.2m/s2)t;
v2 = -(4.6m/s) + (2.2m/s2)(2.0s) = -0.2m/s (toward -x);
v3 = -(4.6m/s) + (2.2m/s2)(3.0s) = +2.0m/s (toward +x).
3. We find the time for the outgoing 200km from
t1 = d1/v1 = (200km)/(90km/h) = 2.22h.
We find the time for the return 200km from
t2 = d2/v2 = (200km)/(50km/h) = 4.00h.
We find the average speed from
average speed = (d1 + d2)/(t1 + tlunch + t2) = (200km + 200km)/(2.22h + 1.00h + 4.00h) = 55km/h.
Because the trip finishes at the starting point, there is no displacement; thus the average velocity is
vav = Dx/Dt = 0
4. We find the average acceleration from
aav = Dv/Dt = [(95km/h)((1h/3.6ks) - 0]/(6.2s) = 4.3m/s2.
5. The position is given by x = At + 6Bt3.
(a) All terms must give the same units, so we have
A ~ x/t = m/s; and B ~ x/t3 = m/s3.
(b) We find the velocity and acceleration by differentiating:
v = dx/dt = A + 18Bt2;
a = dv/dt = 36Bt.
(c) For the given time we have
v = dx/dt = A + 18Bt2 = A + 18B(5.0s)2 = A + (450s2)B;
a = dv/dt = 36Bt =36B(5.0s) = (180s)B.
(d) We find the velocity by differentiating:
v = dx/dt = A - 3Bt-4.
6. We find the average acceleration from
v2 = vo2 + 2aav(x2 - x1);
(11.5m/s)2 = 0 + 2aav(15.0m), which gives aav =4.41m/s2
We find the time required from
x = 1/2(v + vo)t;
15.0m = 1/2(11.5m/s + )), which gives t = 2.61s.
7. We find the average acceleration from
v2 = vo2 + 2aav(x2 - x1);
0 = [(95km/h)(3.6ks/h)]2 + 2aav(0.80m), which gives aav = -4.4 x 102m/s2.
The number of g's is
| aav | = (4.4 x 102m/s2)/[(9.80m/s2)/g] = 44g.
(95km/h)/(3.6ks/h) = 26.4m/s.
(140km/h)(3.6ks/h) = 38.9m/s.
We use a coordinate system with the origin where the motorist passes the police officer.
The location of the speeding motorist is given by
xm = xo + vmt = 0 + (38.9m/s)t.
The location of the police officer is given by
xp = xo + vop(1.00s) + vop(t - 1.00s) + 1/2ap(t< - 1.00s)2
= 0 + (2.64m/s)t + 1/2(2.00m/s2)(t - 1.00s)2.
The officer will reach the speeder when these locations coincide, so we have
xm = xp;
(38.9m/s)t = (26.4m/s)t + 1/2(2.00m/s2)(t - 1.00s)2.
The solution to this quadratic equation are 0.07s and 14.4s.
Because the time must be greater than 1.00s, the result is t = 14.4s.
9. We use a coordinate system with the origin at the ground and up positive.
We can find the initial velocity from the maximum height (where the velocity is zero):
v2 = vo2 + 2ah;
0 = vo2 + 2(-9.80m/s2)(2.55m), which gives vo = 7.07m/s.
When the kangaroo returns to the ground, its displacement is zero. For the entire jump we have
y = yo + vot + 1/2at2;
0 = 0 + (7.07m/s)t + 1/2(-9.80m/s2)t2,
Which gives t = 0 (when the kangaroo jumps), and t = 1.44s.
10. We use a coordinate system with the origin at the ground and up positive.
(a) We find the velocity from
v2 + 2a(y - yo);
v2 = (23.0m/s)2 + 2(-9.8m/s2)(12.0m - 0), which gives v = +17.1m/s
The stone reaches this height on the way up ( the positive sign) and on the way down (the negative sign).
(b) We find the time to reach the height from
v = vo + at;
+17.1m/s = 23.0m/s + (-9.8m/s2)t, which gives t = 0.602s, 4.09s.
(c) there are two answers because the stone reaches this height on the way up (t = 0.602s) and on the way down (t = 4.09s).