Question

A car moved at a constant velocity during the first hour. It stopped for 2 hours at a mall and then moved ahead again at a constant velocity for the next 3 hours. The car finally returned to its starting point with a constant velocity in the next 2.5 hours. Which of the following graphs best represents the car's motion?
First straight line joins ordered pairs 0, 0 and 1, 60, second straight line joins 1, 60 and 3, 60, third straight line joins 3, 60 and 6, 30 and fourth straight line joins 6, 30 and 8.5, 0
First straight line joins ordered pairs 0, 0 and 1, 60, second straight line joins 1, 60 and 3, 60, third straight line joins 3, 60 and 6, 100 and fourth straight line joins 6, 100 and 8.5, 0
First straight line joins ordered pairs 0, 0 and 1, 60, second straight line joins 1, 60 and 3, 60, third straight line joins 3, 60 and 6, 100 and fourth straight line joins 6, 100 and 9.5, 0
First straight line joins ordered pairs 0, 0 and 1, 60, second straight line joins 1, 60 and 6, 100, and third straight line joins 6, 100 and 8.5, 0

Answer 1

You are given a car moved at a constant velocity during the first hour. It stopped for 2 hours at a mall and then moved ahead again at a constant velocity for the next 3 hours. Then you are given the car that has finally returned to its starting point with a constant velocity in the next 2.5 hours. The graph that best represents the car's motion is First straight line joins ordered pairs 0, 0 and 1, 60, second straight line joins 1, 60 and 3, 60, third straight line joins 3, 60 and 6, 100 and fourth straight line joins 6, 100 and 8.5, 0.

Answer 2

Answer:

First straight line joins ordered pairs 0, 0 and 1, 60; second straight line joins 1, 60 and 3, 60; third straight line joins 3, 60 and 6, 100; and fourth straight line joins 6, 100 and 8.5, 0.

Step-by-step explanation:

Starting at home is starting at (0, 0).

Traveling at the same velocity for an hour will connect (0, 0) to (1, 60).

Stopping at the mall for 2 hours will connect (1, 60) to (3, 60).

Going ahead again at a constant velocity for 3 hours will connect (3, 60) to (6, 100).

Returning to the starting point for 2.5 hours will connect (6, 100) to (8.5, 0).