1 (a) The position vectors:
r1=−3i,r2=+7i. 1 (b) The displacement in 8 minutes:
Δx8=7−(−3)=10 km. This took 8 minutes, so, the velocity is
v=t8Δx8=5/4 km/min.
Assuming constant velocity, at t=6 min we have
Δx6=vt6=45⋅6=7.5 km. 2 (a) The particle passes the origin when the origin is 0:
x(t)=0=5.0−2.0t,t=2.5 s.2 (b) The displacement between 4 and 8 s:
Δx=x(8)−x(4),Δx=[5−2⋅8]−[5−2⋅4]=−8 m, which means that at 8 s the car is further from the origin than at 4 s.
3 (a) The velocities are the derivatives of displacement over time:
v(t)=x′(t)=8−4t,v(2)=0,v(3)=−4 m/s.3 (b) The instantaneous speeds are equal to the instantaneous velocities in magnitude:
s(2)=0,s(3)=4 m/s.3 (c) The average velocity is displacement over time (1 s):
x(2)=8⋅2−2⋅22=8 m,x(3)=6 m,Δx=−2 m,vav=tΔx=−2 m/s.
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