Can you answer this problem? a. At what time(s) does the car come to a rest?b. During what interval(s) of time is the car moving to the right?c. During what interval(s) of time is the car speeding up?

Answered: a. At what time(s) does the car come to…

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Question

Can you answer this problem?

a. At what time(s) does the car come to a rest?
b. During what interval(s) of time is the car moving to the right?
c. During what interval(s) of time is the car speeding up?

Expert Solution

Step 1

Since we only answer up to 3 sub-parts, we’ll answer the first 3. Please resubmit the question and specify the other subparts (up to 3) you’d like answered.

In kinematics, the velocity of a particle is defined as the rate of change of displacement. It is a vector quantity.

$\vec{v} = \frac{d \vec{s}}{d t}$

Acceleration of the particle is defined as the rate of change of velocity

$\vec{a} = \frac{d \vec{v}}{d t}$

Step 2

The velocity function of the car is given by

$v (t) = \{\begin{cases} 3 t^{2} - 12 t + 9 & f o r 0 \leq t \leq 3 \\ 6 t - 18 & f o r 3 \leq t \leq 5 \end{cases}$

a. The time at which the car comes to rest is the time at which the velocity is equal t zero. This gives

$3 t^{2} - 12 t + 9 = 0 \Rightarrow 3 t^{2} - 9 t - 3 t + 9 = 0 \Rightarrow 3 t (t - 3) - 3 (t - 3) = 0 \Rightarrow (t - 3) (3 t - 3) = 0 \Rightarrow t = 1 o r 3$

From the second equation we have

$6 t - 18 = 0 \Rightarrow t = 3$

This gives that the car comes to rest at time $1 s$ and $3 s$ .

b. The car is moving towards the right if the velocity is positive and the left if the velocity is negative. Let us calculate the velocity at various time

For $t = 0$ we have $v (0) = 9 m / s$

For $t = 1 s$ we have $v (1) = 3 \times 1^{2} - 12 \times 1 + 9 = 0 m / s$

For $t = 2 s$ we have $v (2) = 3 \times 2^{2} - 12 \times 2 + 9 = - 3 m / s$

For $t = 3 s$ we have $v (3) = 3 \times 3^{2} - 12 \times 3 + 9 = 0 m / s$

For $t = 4 s$ we have $v (4) = 6 \times 4 - 18 = 6 m / s$

For $t = 5 s$ we have $v (5) = 6 \times 5 - 18 = 12 m / s$

Therefore from above, we can see that the car is moving to the right for the intervals $0 \leq t \leq 1$ and $3 \leq t \leq 5$ .

c. The intervals where it is speeding is the time interval for which the velocity is increasing. From the previous answer, we can see that the velocity of the cart is increasing in the intervals $1 \leq t \leq 2$ and $3 \leq t \leq 5$ intervals.

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Solved in 3 steps

a. At what time(s) does the car come to a rest? b. During what interval(s) of time is the car moving to the right? c. During what interval(s) of time is the car speeding up?