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 x = v tc. (a) During the first part of the motion, the displacement is x1 = 40
9. The values used in the problem statement make it easy to see that the first part of the trip (at 100 km/h) takes 1 hour, and the second part (at 40 km/h) also takes 1 hour. Expressed in decimal form, the time left is 1.25 hour, and the distance that re
19. We represent its initial direction of motion as the +x direction, so that v0 = +18 m/s and v = 30 m/s (when t = 2.4 s). Using Eq. 2-7 (or Eq. 2-11, suitably interpreted) we find aavg = (-30 m/s) - (+1 m/s) = - 20 m/s 2 2.4 s
which indicates that the a
15. We use Eq. 2-4. to solve the problem. (a) The velocity of the particle is v= dx d = (4 - 12t + 3t 2 ) = -12 + 6t . dt dt
Thus, at t = 1 s, the velocity is v = (12 + (6)(1) = 6 m/s. (b) Since v < 0, it is moving in the negative x direction at t = 1 s.
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,