a) The average acceleration is [tex]0.5 m/s^2[/tex]
b) Instantaneous acceleration: 0 at t = 0, [tex]1.0 m/s^2[/tex] at t = 5.00 s
c) Find the two graphs in attachment
Explanation:
a)
The car velocity as a function of time is given by
[tex]v(t)=\alpha + \beta t^2[/tex]
where
[tex]\alpha = 3.00 m/s\\\beta = 0.100 m/s^3[/tex]
In order to calculate the average acceleration, we have to calculate the velocity at t = 0 and at t = 5.00 s, then calculate the change in velocity, and then divide by the time interval.
The velocity at t = 0 is:
[tex]v(0)=\alpha + \beta (0)^2 = \alpha = 3.00 m/s[/tex]
The velocity at t = 5.00 s is:
[tex]v(5)=\alpha + \beta (5)^2=3.00 + 0.1 \cdot 25 =5.50 m/s[/tex]
So the change in velocity is
[tex]\Delta v = v(5)-v(0)=5.50-3.00 = 2.50 m/s[/tex]
And since the time interval is 5.00 s, the average acceleration is
[tex]a=\frac{\Delta v}{\Delta t}=\frac{2.50}{5.00}=0.5 m/s^2[/tex]
b)
In order to find the instantaneous acceleration, we have to calculate the derivative of the velocity function.
Differentiating the velocity, we get:
[tex]a(t)=v'(t)=(\alpha + \beta t^2)'=2\beta t[/tex]
Therefore, the instantaneous acceleration at t = 0 is
[tex]a(0)=2\beta \cdot 0 = 0[/tex]
While the instantaneous acceleration at t = 5.00 s is
[tex]a(5)=2\beta \cdot 5 = 1.00 m/s^2[/tex]
c)
Find in attachment the velocity-time and acceleration-time graph. We observe that:
- For the velocity-time graph, the curve is a parabola, since the velocity is proportional to [tex]t^2[/tex]. The initial velocity at t = 0 is [tex]v=\alpha[/tex], then it increases up to v = 5.50 m/s when t = 5.00 s
- For the acceleration-time graph, we observe that the curve is a straight line passing through the origin, because [tex]\alpha(t)=2\beta t[/tex], so the acceleration is directly proportional to t. It starts from zero when t = 0 and it becomes equal to [tex]a=1.0 m/s^2[/tex] when t = 5.00 s.
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