55. A dog starts at the origin and runs forward at 6 m/s for 1.5 s and then turns around to fetch the ball by
running backward at 7 m/s for 3 s. If the dog runs
back to the origin at 4 m/s, then how much time has
elapsed between the start and when he returns with
the ball?
0-
Position vs. time
61
-5
-10
Position (m)
-15
-20
-250
0.2
0.8
1.0
1.2
0.4 0.6
Time (s)

7.5 seconds

Explanation:

Given that a dog starts at the origin and runs forward at 6 m/s for 1.5 s

From speed formula

Speed = distance/time

Substitute speed and time into the formula

6 = distance/1.5

Distance = 6 × 1.5 = 9 m

and then turns around to fetch the ball by

running backward at 7 m/s for 3 s. If the dog runs.

Calculate the distance covered

Distance = 7 × 3 = 21 m

This means that the dog ran beyond the original position.

The elapsed distance = 21 - 9 = 12m

back to the origin at 4 m/s, by covering distance 12 m

Calculate the time.

Speed = distance/time

Time = 12/4 = 3s

The time elapsed between the start and when he returns with the ball will be:

T = 1.5 + 3 × 3

T = 7.5 seconds

d is the correct answer