A dog runs 100 m away from its master in a straight line in 8.4 s, and then runs halfway back in 1/3 the time.

Question

Calculate:

a) its average speed

b) its average velocity

Answer 1

a. Avg. speed = 150 m / (8.4 + 2.8)s

b. Avg velocity =  50 m / 11.2 s

Answer 2

S1=100m/8.4s

S1=11.9m/s

S2=50/(8.4s/3)

S2=17.86m/s

Average Speed=(17.86+11.9)/2

=14.88m/s

V=displacement/time

V=50m/11.2s

V=4.46m/s

Answer 3

Total distance traveled = 100 + (1/2 * 100) = 150 m

Time taken = 8.4s + (1/3 * 8.4) = 11.2 s

Average speed = Total distance / total time = 13.4 m/s

Total displacement = 100 + (-50) = 50 m

Time taken = 11.2 s

Average velocity = Total disp. / Total time = 4.5 m/s

Hope this helps.

[e-mail]

Answer 4

when the dog runs away from its master, its velocity is

  v1 = 100 / 8.4

  v1 = 11.9 m/s.

when he runs halfway back , its velocity is

  v2 = 50 / 2.8

  v2 = 17.85 m/s

its average speed is( v1 = v2 ) /2

                              14.875 m/s

as the distance is a scalar quantity therefore all the travelling will be counted.

the average velocity requires onl displacement

therefore

the displacement, the net distance, = 100 - 50 = 50 m

and the time for this net displacement will be the whole time spent during the whole journey

therefore  time is 8.4 s + 2.8 s = 11.2 s .



average velocity is = 50 / 11.2

                              4.46 m/s away from the master