A truck travels due east for a distance of 1.9 km, turns around and goes due west for 9.3 km, and finally turns around again and travels 3.4 km due east. What is the total distance that the truck travels? What are the magnitude and direction of the truck's displacement?

The total distance of the truck is 14.6 km

The magnitude and direction of truck's disaplacement is 4 km east

Explanation:

Distance is the total path covered by the trcuk regardless of the direction since distance is described by magnitude only.

The total distance traveled by the truck is given by;

Total Distance = 1.9 km + 9.3 km + 3.4 km

Total Distance = 14.6 km

Displacement is the change in postion at a time interval. In dispalcement direction is considered.

let eastward motion be positive

let westward motion be negative

Displacement = 1.9km east - 9.3 km west + 3.4 km east

Displacement = -4 km east

Therefore, the magnitude and direction of dispalcement = 4 km east

The uniform motion ratios allow finding which intervals have constant velocity, the correct answers are

         BC       Positive velocity

         CD       Zero speed

         DE       Negative velocity

Kinematics studies the movement of bodies, establishing relationships between position, velocity and acceleration. In the special case that the acceleration is zero the motion is called uniform movement

        v =

        Δx = v Δt

Where v is average velocity, Δx is the position variation and Δt is the time variation in the analyzed interval.

This equation we see that the relationship between displacement and time is linear, let's analyze each interval given in the graph

AB interval

   In this interval the graph shows a curve whereby the movement is accelerated

BC interval

  In this interval the graph is a line with a positive slope, therefore the movement is uniform with positive velocity

CD interval

    In this part we have a line of zero slope

     In motion it is uniform with zero speed, that is, the car is stopped

DE interval

    The graph of this section is a line with a negative slope, consequently the velocity is negative.

With this analysis we see that the only interval that the movement is accelerated is the AB interval  

In conclusion, using the uniform motion relationships we can find that intervals have constant velocity, the correct answers are

         BC     Positive velocity

         CD     Zero speed  

         DE     Negative velocity

Learn more about average speed here:

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