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.
s
b.) when is the velocity of the car negative?
from s to s.
c.) when is the acceleration of the car zero?
from s to s.
d.) what is the acceleration of the car at t = 0.9 s?
$m/s^{2}$
e.) what is the acceleration of the car at t = 16.4 s?
$m/s^{2}$
f.) when is the car slowing down?
from s to s.
g.) what is the maximum speed the car reaches during the time shown?
$m/s$

Explanation:

Step1: Analyze the velocity - time graph

• For part a), look for the point where the velocity \(v = 0\).
• For part b), identify the time interval where \(v<0\).
• For part c), acceleration \(a=\frac{\Delta v}{\Delta t}\). When \(a = 0\), \(\Delta v=0\) (constant - velocity region).
• For part d), use the formula \(a=\frac{v - u}{t}\) for the first - stage (from \(t = 0\) to \(t=6\) s, \(u = 30\) m/s, \(v = 60\) m/s).
• For part e), use the formula \(a=\frac{v - u}{t}\) for the third - stage (from \(t = 12\) s to \(t = 30\) s, \(u = 60\) m/s, \(v=-20\) m/s).
• For part f), the car is slowing down when \(a\) and \(v\) have opposite signs.
• For part g), the maximum speed is the highest value of \(|v|\) on the graph.

Step2: Calculate each part

• a)

We look for the \(t\) - value when \(v = 0\). By observing the graph, when \(v = 0\), \(t = 27\) s.

• b)

The velocity is negative when \(v<0\). From the graph, \(v<0\) from \(t = 27\) s to \(t = 30\) s.

• c)

Acceleration \(a=\frac{\Delta v}{\Delta t}\). When \(a = 0\), \(\Delta v = 0\). This occurs from \(t=6\) s to \(t = 12\) s (constant - velocity region, \(v = 60\) m/s).

• d)

For \(t\in[0,6]\) s, \(u = 30\) m/s, \(v = 60\) m/s, \(t = 6\) s. Using \(a=\frac{v - u}{t}\), \(a=\frac{60 - 30}{6}=\frac{30}{6}=5\) m/s². At \(t = 0.9\) s (which is in the interval \(t\in[0,6]\) s), \(a = 5\) m/s².

• e)

For \(t\in[12,30]\) s, \(u = 60\) m/s, \(v=-20\) m/s, \(t=30 - 12=18\) s. Using \(a=\frac{v - u}{t}\), \(a=\frac{-20 - 60}{18}=\frac{-80}{18}\approx - 4.44\) m/s². At \(t = 16.4\) s (which is in the interval \(t\in[12,30]\) s), \(a\approx - 4.44\) m/s².

• f)

The car is slowing down when \(a\) and \(v\) have opposite signs.

• In the first stage (\(t\in[0,6]\) s), \(a>0\) and \(v>0\) (speeding up).
• In the second stage (\(t\in[6,12]\) s), \(a = 0\) (constant speed).
• In the third stage (\(t\in[12,27]\) s), \(a<0\) and \(v>0\) (slowing down).

So the car is slowing down from \(t = 12\) s to \(t = 27\) s.

• g)

The maximum speed is the highest value of \(|v|\). From the graph, \(|v|_{max}=60\) m/s.

Answer:

a. \(27\)
b. \(27\); \(30\)
c. \(6\); \(12\)
d. \(5\)
e. \(-\frac{40}{9}\approx - 4.44\)
f. \(12\); \(27\)
g. \(60\)