Question

A car’s velocity as a function of time is given by vx(t) = α + βt2, where α = 3.00 m/s and β = 0.100 m/s3. (a) Calculate the average acceleration for the time interval t = 0 to t = 5.00 s. (b) Calculate the instantaneous acceleration for t = 0 and t = 5.00 s. (c) Draw vx-t and ax-t graphs for the car’s motion between t = 0 and t = 5.00 s.

Answer 1

a) The average acceleration is $0.5 m/s^2$

b) Instantaneous acceleration: 0 at t = 0, $1.0 m/s^2$ 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

$v(t)=\alpha + \beta t^2$

where

$\alpha = 3.00 m/s\\\beta = 0.100 m/s^3$

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:

$v(0)=\alpha + \beta (0)^2 = \alpha = 3.00 m/s$

The velocity at t = 5.00 s is:

$v(5)=\alpha + \beta (5)^2=3.00 + 0.1 \cdot 25 =5.50 m/s$

So the change in velocity is

$\Delta v = v(5)-v(0)=5.50-3.00 = 2.50 m/s$

And since the time interval is 5.00 s, the average acceleration is

$a=\frac{\Delta v}{\Delta t}=\frac{2.50}{5.00}=0.5 m/s^2$

b)

In order to find the instantaneous acceleration, we have to calculate the derivative of the velocity function.

Differentiating the velocity, we get:

$a(t)=v'(t)=(\alpha + \beta t^2)'=2\beta t$

Therefore, the instantaneous acceleration at t = 0 is

$a(0)=2\beta \cdot 0 = 0$

While the instantaneous acceleration at t = 5.00 s is

$a(5)=2\beta \cdot 5 = 1.00 m/s^2$

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 $t^2$. The initial velocity at t = 0 is $v=\alpha$, 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 $\alpha(t)=2\beta t$, so the acceleration is directly proportional to t. It starts from zero when t = 0 and it becomes equal to $a=1.0 m/s^2$ when t = 5.00 s.

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