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.<

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answered: 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.

answered: 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,

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