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17. An object's velocity can be graphed as a function of time as seen on the right. From t=2s to t=4 s, the velocity of
the object is 3 m/s.

17 An Objects Velocity Can Be Graphed As A Function Of Time As Seen On The Right From T2s To T4 S The Velocity Of The Object Is 3 Ms class=

Sagot :

semsee45

Answer:

(a)  v = 0 m/s

(b)  v = 6 m/s

(c)  6 m

(d)  a = 0 m/s²

(e)  a = 0.75 m/s²

Explanation:

A velocity-time graph shows the velocity (speed) and direction an object travels over a specific period of time.

  • y-axis = velocity (in m/s).
  • x-axis = time (in seconds).

A horizontal line means constant velocity.

When v = 0 m/s, the object is at rest.

Acceleration is the slope of the line. (A positive slope is acceleration, and a negative slope is deceleration).

Displacement (distance traveled) is the area under the graph.

Part (a)

The object's initial velocity is when t = 0 s.

Therefore, from inspection of the graph, the initial velocity is:

  • v = 0 m/s.

Part (b)

The object's final velocity is when t = 8 s.

Therefore, from inspection of the graph, the final velocity is:

  • v = 6 m/s

Part (c)

To calculate the displacement of the object from t = 0 s to t = 2 s, find the area under the graph between those times.

The area is a triangle with base 2 and height 3.  Therefore, using the area of a triangle formula:

  • [tex]\sf Displacement=\dfrac{1}{2} \times 2 \times 3=6\;m[/tex]

Part (d)

The line between t = 2 s and t = 4 s is horizontal.  Therefore, the velocity between these times is constant and so the acceleration of the object is zero:

  • a = 0 m/s²

Part (e)

To calculate the acceleration of the object from t = 4 s to t = 8 s, find the slope of the line between these two points:

[tex]\implies \textsf{slope}=\dfrac{\textsf{change in $y$}}{\textsf{change in $x$}}=\sf \dfrac{6-3}{8-4}=\dfrac{3}{4}=0.75[/tex]

Therefore, the acceleration of the object from t = 4 s to t = 8 s is:

  • a = 0.75 m/s²
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