Mathematics, 27.08.2019 06:00 yani246

Determine the magnitude of the acceleration for the speeding up and slowing down phase.

Determine the magnitude of the acceleration for the speeding up and slowing down phase.<

Answers: 1

Show answers

Answers

Answer from: snoopyqueen2597

The magnitude of the acceleration for the speeding up phase is $\boxed{ \ \boxed{a = 15 \ m/s^2} \ }$ The magnitude of the acceleration for the slowing down phase is $\boxed{ \ \boxed{a = - 5 \ m/s^2} \ }$ Further explanation

From the v-t graph, we see that:

the object at rest at time interval 0 ≤ t < 0.20 and 0.40 ≤ t ≤ 0.60 (in seconds)the object is speeding up and slowing down at interval 0.20 < t 0.40 (in seconds)

In physics, acceleration is the rate of change of velocity per unit time.

$\boxed{\boxed{ \ a = \frac{\Delta v}{\Delta t} \ in \ m/s^2}}$

In mathematics, acceleration is the gradient or slope of the line with the vertical axis is velocity (v) and the horizontal axis is time (t).

Remember this, $\boxed{ \ gradient \ m = \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1} \ }$

The same analogy for the acceleration formula, i.e.

$\boxed{\boxed{ \ a = \frac{v_2 - v_1}{t_2 - t_1} \ in \ m/s^2}}$

The two points that cause the upward-sloping line are

$\boxed{v_1 = 0 \ m/s} \ \boxed{t_1 = 0.20 \ s}$

$\boxed{v_2 = 0.75 \ m/s} \ \boxed{t_2 = 0.25 \ s}$

$\boxed{ \ a = \frac{0.75 - 0}{0.25 - 0.20}}$

$\boxed{ \ a = \frac{0.75}{0.05}}$

We get the magnitude of the acceleration for the speeding up phase is

$\boxed{ \ \boxed{a = 15 \ m/s^2} \ }$

The two points that cause the downward-sloping line are

$\boxed{v_1 = 0.75 \ m/s} \ \boxed{t_1 = 0.25 \ s}$

$\boxed{v_2 = 0 \ m/s} \ \boxed{t_2 = 0.40 \ s}$

$\boxed{ \ a = \frac{0 - 0.75}{0.40 - 0.25}}$

$\boxed{ \ a = \frac{- 0.75}{0.15}}$

We get the magnitude of the acceleration for the speeding up phase is

$\boxed{ \ \boxed{a = - 5 \ m/s^2} \ }$ with a negative sign.

Note:

Positive slope, in other words speeding up, produces a positive sign of acceleration. The acceleration is in a similar direction as the velocity. Example: free-falling object.Negative slope, in other words slowing down, produces a negative sign of acceleration. The acceleration is precisely in the opposite direction as the velocity. Example: the car is slowing or braking.Acceleration is precisely a vector quantity defined as the rate at which an object changes its velocity.Learn moreInterpreting the chart that shows the time, initial velocity, and final velocity linkDescribing different velocity-time graph lines linkDefining the acceleration link

Keywords: determine, the magnitude, acceleration, speeding up, slowing down, phase, time interval, at rest, the rate, change, velocity, unit, upward, downward, sloping lines, gradient

Determine the magnitude of the acceleration for the speeding up and slowing down phase.

Answer from: moneytt7508

The magnitude of the acceleration for speeding up phase is $\boxed{15m/{s^2}}$ and for slowing down phase is $\boxed{ - 5m/{s^2}}$

Further explanation

Acceleration can be stated as the rate of change of velocity with respect to time in a specified direction. It is denoted as

Velocity can be stated as the rate of change of distance with respect to time in a specified direction. It is denoted as .

Both the terms are vector quantity.

The magnitude of the acceleration can be calculated as,

$\begin{aligned}a&= \frac{{\Delta v}}{{\Delta t}}\\&= \frac{{{v_2} - {v_1}}}{{{t_2} - {t_1}}}\\\end{aligned}$

Here, ${v_1}{\text{ and }}{v_2}$ are the velocities at the time ${t_1}{\text{ and }}{t_2}$ respectively.

Step 1:

The horizontal axis represents time and vertical axis represents

It can be seen from the given graph that the object is on the rest at time interval $0 \leqslant t \leqslant 0.20$ and $0.40 \leqslant t \leqslant 0.60$ .

It can be observed from the given graph the velocity function is rising in the interval of time $0.20{\text{ to }}0.30$ .

The velocity of the object at time $0.20{\text{ s}}$ is 0m/s and the velocity of the object at time $0.25{\text{ s}}$ is 0.75m/s .

Therefore, the values of ${v_1},{\text{ }}{v_2},{\text{ }}{t_1}{\text{ and }}{t_2}$ can be written as,

${v_1} = 0m/s,{\text{ }}{t_1} = 0.20s,{\text{ }}{v_2} = 0.75m/s{\text{ and }}{t_2} = 0.25s$

Substitute the value of ${v_1},{\text{ }}{v_2},{\text{ }}{t_1}{\text{ and }}{t_2}$ in the formula of acceleration to find the value of acceleration as,

$\begin{aligned}a&= \frac{{{v_2} - {v_1}}}{{{t_2} - {t_1}}}\\&= \frac{{0.75 - 0}}{{0.25 - 0.20}}\\&= \frac{{0.75}}{{0.05}}\\&= 15m/{s^2}\\\end{aligned}$

The magnitude of the acceleration for the speeding up phase is $15m/{s^2}$ .

Step 2:

It can be observed from the given graph the velocity function is declining in the interval of time $0.30{\text{ to }}0.40$ .

The velocity of the object at time $0.25{\text{ s}}$ is ${\text{0}}{\text{.75m/s}}$ and the velocity of the object at time $0.40 \,\text{s}$ is ${\text{0m/s}}$ .

Therefore, the values of ${v_1},{\text{ }}{v_2},{\text{ }}{t_1}{\text{ and }}{t_2}$ can be written as,

${v_1} = 0.75m/s,{\text{ }}{t_1} = 0.25s,{\text{ }}{v_2} = 0m/s{\text{ and }}{t_2} = 0.40s$

Substitute the value of ${v_1},{\text{ }}{v_2},{\text{ }}{t_1}{\text{ and }}{t_2}$ in the formula of acceleration to find the value of acceleration as,

$\begin{aligned}a&= \frac{{{v_2} - {v_1}}}{{{t_2} - {t_1}}}\\&= \frac{{0 - 0.75}}{{0.40 - 0.25}}\\&= \frac{{ - 0.75}}{{0.15}} \\ &= - 5m/{s^2} \\\end{aligned}$

Here, the magnitude of the acceleration for the slowing down phase is negative.

Note:

Negative slope represents the slowing down phase and it is in the downward direction.

Example: when the car is slowing down its speed or applying brakes.

Positive slope represents the speeding up phase and it is in the upward direction.

Example: free falling object

Learn more:

Learn more about the function is graphed below link Learn more about the symmetry for a function link Learn more about midpoint of the segment link

Answer details:

Grade: Middle school

Subject: Mathematics

Chapter: Speed and distance

Keywords: Speed, velocity, distance, acceleration, vector quantity, slowing down, braking, change, magnitude, unit, upward, slope, time interval

Answer from: Quest

hc ghfrfgb

step-by-step explanation:

gedfgb nbgfrezrdfgbn bgfderdfgbvn bvfdsvcb bvfdfgb gfdgb gfr

Answer from: Quest

find the vertical and horizontal asymptotes of the function f x = 5 x − 1 . to find the vertical asymptote, equate the denominator to zero and solve for . since the degree of the polynomial in the numerator is less than that of the denominator, the horizontal asymptote is .

step-by-step explanation:

Another question on Mathematics

Mathematics, 21.06.2019 19:00

What are the solutions of the system? solve by graphing. y = -x^2 -6x - 7 y = 2

Answers: 2

Answer

Mathematics, 21.06.2019 21:00

Adesigner charges a one time fee of $200 plus $40 an hour for each project. write an expression to represent how much money the designer will make for a project

Answers: 1

Answer

Mathematics, 21.06.2019 22:30

Nicole is making carrot bread she needs 5 cups of carrots for 2 cups of sugar.what is the ratio of cups of sugar to cups of carrots that nicole will need? write the ration in three different ways

Answers: 1

Answer

Mathematics, 22.06.2019 01:00

Astudent drew a circle and two secant segment. he concluded that if pq ~= ps, then qr ~= st. do you agree with the student’s conclusion? why or why not?

Answers: 1

Answer

You know the right answer?

Determine the magnitude of the acceleration for the speeding up and slowing down phase.
...

Questions