You drive a beat-up pickup truck along a straight road for 8.4 km at 70 km/h, at which point the truck runs out of gasoline and stops. Over the next 30 minutes, you walk another 2.0 km farther along the road to a gasoline station.

a) 10.4km

b) 0.62h

C) 17km/h

d) 9.1km/h

Explanation:

A) suppose you move in the positive direction of an x axis, from a first position of x1 = 0 to a second position of x2 at the station. That second position must be at x2 = 8.4 km + 2.0 km = 10.4 km.

B) time interval Dtwlk (= 0.50 h), but we don't know the driving time interval Dt/dr. However, we know that for the drive the displacement Dx/dr is 8.4 km and the average velocity vavg,dr is 70 km/h.

This average velocity is the ratio of the displacement for the drive to the time interval for the drive:

dv = dx/ dt

dt = dx/dv = 8.4/70 = 0.12h

t = ti + two

t = 0.12 + 0.50 = 0.62h

C) avg speed for the entire trip is the ratio of the displacement of 10.4 km for the entire trip to the time interval of 0.62 h for the entire trip. 

Avg v = distance/time

10.4/0.62 = 16.8km/h

D) average speed is the ratio of the total distance you move to the total time interval you take to make that move. The total distance is 8.4 km + 2.0 km + 2.0 km = 12.4 km. The total time interval is 0.12 h + 0.50 h + 0.75 h = 1.37 h.

Avg speed = 12.4/1.37 = 9.1km/h

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j