From the v-t graph, we see that:
In physics, acceleration is the rate of change of velocity per unit time.
[tex]\boxed{\boxed{ \ a = \frac{\Delta v}{\Delta t} \ in \ m/s^2}}[/tex]
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, [tex]\boxed{ \ gradient \ m = \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1} \ }[/tex]
The same analogy for the acceleration formula, i.e.
[tex]\boxed{\boxed{ \ a = \frac{v_2 - v_1}{t_2 - t_1} \ in \ m/s^2}}[/tex]
The two points that cause the upward-sloping line are
[tex]\boxed{v_1 = 0 \ m/s} \ \boxed{t_1 = 0.20 \ s}[/tex]
[tex]\boxed{v_2 = 0.75 \ m/s} \ \boxed{t_2 = 0.25 \ s}[/tex]
[tex]\boxed{ \ a = \frac{0.75 - 0}{0.25 - 0.20}} [/tex]
[tex]\boxed{ \ a = \frac{0.75}{0.05}} [/tex]
We get the magnitude of the acceleration for the speeding up phase is
[tex]\boxed{ \ \boxed{a = 15 \ m/s^2} \ }[/tex]
The two points that cause the downward-sloping line are
[tex]\boxed{v_1 = 0.75 \ m/s} \ \boxed{t_1 = 0.25 \ s}[/tex]
[tex]\boxed{v_2 = 0 \ m/s} \ \boxed{t_2 = 0.40 \ s}[/tex]
[tex]\boxed{ \ a = \frac{0 - 0.75}{0.40 - 0.25}} [/tex]
[tex]\boxed{ \ a = \frac{- 0.75}{0.15}} [/tex]
We get the magnitude of the acceleration for the speeding up phase is
[tex]\boxed{ \ \boxed{a = - 5 \ m/s^2} \ }[/tex] with a negative sign.
Note:
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The magnitude of the acceleration for speeding up phase is [tex]\boxed{15m/{s^2}}[/tex] and for slowing down phase is [tex]\boxed{ - 5m/{s^2}}[/tex]
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 [tex]a[/tex]
Velocity can be stated as the rate of change of distance with respect to time in a specified direction. It is denoted as [tex]v[/tex].
Both the terms are vector quantity.
The magnitude of the acceleration can be calculated as,
[tex]\begin{aligned}a&= \frac{{\Delta v}}{{\Delta t}}\\&= \frac{{{v_2} - {v_1}}}{{{t_2} - {t_1}}}\\\end{aligned}[/tex]
Here, [tex]{v_1}{\text{ and }}{v_2}[/tex] are the velocities at the time [tex]{t_1}{\text{ and }}{t_2}[/tex] 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 [tex]0 \leqslant t \leqslant 0.20[/tex] and [tex]0.40 \leqslant t \leqslant 0.60[/tex].
It can be observed from the given graph the velocity function is rising in the interval of time [tex]0.20{\text{ to }}0.30[/tex].
The velocity of the object at time [tex]0.20{\text{ s}}[/tex] is [tex]0m/s[/tex] and the velocity of the object at time [tex]0.25{\text{ s}}[/tex] is [tex]0.75m/s[/tex].
Therefore, the values of [tex]{v_1},{\text{ }}{v_2},{\text{ }}{t_1}{\text{ and }}{t_2}[/tex] can be written as,
[tex]{v_1} = 0m/s,{\text{ }}{t_1} = 0.20s,{\text{ }}{v_2} = 0.75m/s{\text{ and }}{t_2} = 0.25s[/tex]
Substitute the value of [tex]{v_1},{\text{ }}{v_2},{\text{ }}{t_1}{\text{ and }}{t_2}[/tex] in the formula of acceleration to find the value of acceleration as,
[tex]\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}[/tex]
The magnitude of the acceleration for the speeding up phase is [tex]15m/{s^2}[/tex].
Step 2:
It can be observed from the given graph the velocity function is declining in the interval of time [tex]0.30{\text{ to }}0.40[/tex].
The velocity of the object at time [tex]0.25{\text{ s}}[/tex] is [tex]{\text{0}}{\text{.75m/s}}[/tex]and the velocity of the object at time [tex]0.40 \,\text{s}[/tex] is [tex]{\text{0m/s}}[/tex].
Therefore, the values of [tex]{v_1},{\text{ }}{v_2},{\text{ }}{t_1}{\text{ and }}{t_2}[/tex] can be written as,
[tex]{v_1} = 0.75m/s,{\text{ }}{t_1} = 0.25s,{\text{ }}{v_2} = 0m/s{\text{ and }}{t_2} = 0.40s[/tex]
Substitute the value of [tex]{v_1},{\text{ }}{v_2},{\text{ }}{t_1}{\text{ and }}{t_2}[/tex] in the formula of acceleration to find the value of acceleration as,
[tex]\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}[/tex]
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:
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
