Question

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

Answer 1

• 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 more

1. Interpreting the chart that shows the time, initial velocity, and final velocity brainly.com/question/4319751
2. Describing different velocity-time graph lines brainly.com/question/11205819
3. Defining the acceleration brainly.com/question/6753991

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

Answer 2

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 $a$

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

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:  

1. Learn more about the function is graphed below brainly.com/question/9590016
2. Learn more about the symmetry for a function brainly.com/question/1286775
3. Learn more about midpoint of the segment brainly.com/question/3269852

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