Question

the graph below shows the velocity of a car with respect to time. use this graph to answer the following questions.
car velocity vs. time
a.) state the time at which the velocity of the car is zero.
27 × s
b.) when is the velocity of the car negative?
from 27 × s to 30 ✓ s.
c.) when is the acceleration of the car zero?
from 6 ✓ s to 12 ✓ s.
d.) what is the acceleration of the car at t = 0.9 s?
5 ✓ m/s²
e.) what is the acceleration of the car at t = 16.4 s?
-4.44 ✓ m/s²
f.) when is the car slowing down?
from 12 ✓ s to 27 × s.
g.) what is the maximum speed the car reaches during the time shown?
60 ✓ m/s

Explanation:

Step1: Analyze velocity - time graph for zero velocity

On a velocity - time graph, the velocity is zero when the graph crosses the time - axis. Looking at the given graph, we can see that the velocity is zero at \(t = 24\) s.

Step2: Determine when velocity is negative

The velocity is negative when the graph is below the time - axis. From the graph, the velocity is negative from \(t = 27\) s to \(t = 30\) s.

Step3: Find when acceleration is zero

Acceleration is the slope of the velocity - time graph. When the slope is zero (i.e., the graph is a horizontal line), the acceleration is zero. From the graph, the graph is horizontal from \(t = 6\) s to \(t = 12\) s.

Step4: Calculate acceleration at \(t = 0.9\) s

The acceleration \(a=\frac{\Delta v}{\Delta t}\). For the first part of the graph (from \(t = 0\) s to \(t = 6\) s), \(v_1 = 30\) m/s and \(v_2=60\) m/s, \(\Delta t=6\) s. The slope \(a=\frac{60 - 30}{6}=5\) m/s². Since \(t = 0.9\) s is in this interval, the acceleration is \(5\) m/s².

Step5: Calculate acceleration at \(t = 16.4\) s

For the part of the graph from \(t = 12\) s to \(t = 30\) s, \(v_1 = 60\) m/s and \(v_2=- 20\) m/s, \(\Delta t=30 - 12=18\) s. The slope \(a=\frac{-20 - 60}{30 - 12}=\frac{-80}{18}\approx - 4.44\) m/s². Since \(t = 16.4\) s is in this interval, the acceleration is \(-4.44\) m/s².

Step6: Determine when the car is slowing down

The car is slowing down when the velocity and acceleration have opposite signs.

• From \(t = 6\) s to \(t = 12\) s, \(a = 0\), so no slowing down.
• From \(t = 0\) s to \(t = 6\) s, \(a>0\) and \(v>0\) (speeding up).
• From \(t = 12\) s to \(t = 24\) s, \(a<0\) and \(v>0\) (slowing down).
• From \(t = 24\) s to \(t = 30\) s, \(a<0\) and \(v<0\) (speeding up in the negative direction)

So the car is slowing down from \(t = 12\) s to \(t = 24\) s.

Step7: Find maximum speed

The maximum speed is the highest value of the velocity on the graph. From the graph, the maximum speed is \(60\) m/s.

Answer:

a. \(24\) s
b. From \(27\) s to \(30\) s
c. From \(6\) s to \(12\) s
d. \(5\) m/s²
e. \(-4.44\) m/s²
f. From \(12\) s to \(24\) s
g. \(60\) m/s