Question

use the velocity - time graph below to respond to the questions that follow. the graph below represents the motion of an object.
for the first 5 seconds of motion, the objects acceleration is equal to
description of the objects motion during the first 5 seconds of motion
description of the objects motion between a and b.
between b and c, the objects acceleration is equal to
the total distance covered by the object is equal to
drag & drop the answer

Explanation:

Step1: Calculate acceleration for the first 5 seconds

The formula for acceleration \(a=\frac{\Delta v}{\Delta t}\). Here, \(v_{initial} = 0\ m/s\), \(v_{final}=20\ m/s\), \(\Delta t = 5\ s\). So \(a=\frac{20 - 0}{5}=4\ m/s^{2}\)

Step2: Describe motion in first 5 seconds

Since acceleration is constant (\(a = 4\ m/s^{2}>0\)), the object has a positive constant acceleration.

Step3: Describe motion between A and B

The velocity is \(v = 20\ m/s\) (constant). So the object has a positive constant velocity.

Step4: Calculate acceleration between B and C

\(v_{initial}=20\ m/s\), \(v_{final}=0\ m/s\), \(\Delta t=14 - 10 = 4\ s\). Using \(a=\frac{\Delta v}{\Delta t}=\frac{0 - 20}{4}=- 5\ m/s^{2}\)

Step5: Calculate total distance (area under velocity - time graph)

The graph is a trapezoid. Area formula \(A=\frac{(b_1 + b_2)h}{2}\), where \(b_1=10\ s\), \(b_2 = 14\ s\), \(h = 20\ m/s\). \(A=\frac{(10 + 14)\times20}{2}=240\div2 = 190\ m\) (using the sum of areas of triangle, rectangle and triangle: \(\frac{1}{2}\times5\times20+5\times20+\frac{1}{2}\times4\times20=50 + 100+40=190\ m\))

Answer:

For the first 5 seconds of motion, the object's acceleration is equal to \(4\ m/s^{2}\). Description of the object's motion during the first 5 seconds of motion: positive constant acceleration. Description of the object's motion between A and B: Positive Constant Velocity. Between B and C, the object's acceleration is equal to \(-5\ m/s^{2}\). The total distance covered by the object is equal to \(190\ m\)