current attempt in progress\nyour answer is incorrect.\nthe figure below shows the velocity v(t) in ft/sec…

current attempt in progress\nyour answer is incorrect.\nthe figure below shows the velocity v(t) in ft/sec of an object moving on the number line, with positive velocities moving to the right.\nhow far is the object at t = 4 seconds from its starting point, the origin?\nthe object will be feet to the right of the starting point.\netextbook and media

current attempt in progress\nyour answer is incorrect.\nthe figure below shows the velocity v(t) in ft/sec of an object moving on the number line, with positive velocities moving to the right.\nhow far is the object at t = 4 seconds from its starting point, the origin?\nthe object will be feet to the right of the starting point.\netextbook and media

Answer

Explanation:

Step1: Analyze the velocity - time graph

The displacement of an object from the starting - point is given by the net area under the velocity - time graph. The area above the time - axis is positive (right - ward motion) and the area below the time - axis is negative (left - ward motion).

Step2: Calculate the area of the first part

The first part of the graph is a rectangle from (t = 0) to (t = 1) with height (v=35) ft/sec and base (b = 1) sec. The area of a rectangle (A_{1}) is (A_{1}=35\times1 = 35) square feet.

Step3: Calculate the area of the second part

The second part of the graph is a triangle from (t = 1) to (t = 4) with height (h=- 35) ft/sec and base (b = 3) sec. The area of a triangle (A=\frac{1}{2}\times base\times height), so (A_{2}=\frac{1}{2}\times3\times(-35)=-\frac{105}{2}=- 52.5) square feet.

Step4: Calculate the net displacement

The net displacement (D) is the sum of the areas, (D=A_{1}+A_{2}). So (D = 35-52.5=-17.5) feet. The negative sign means the object is to the left of the starting - point. The distance from the starting - point is (|D| = 17.5) feet.

Answer:

17.5