Question

the speed v of an automobile moving on a straight road is given in meters per second as a function of time t in seconds by the following equation: v(t)=4 + 2t^3. 6. what is the acceleration of the automobile at t = 2 s? (a) 12 m/s^2 (b) 16 m/s^2 (c) 20 m/s^2 (d) 24 m/s^2 (e) 28 m/s^2 7. how far has the automobile traveled in the interval between t = 0 and t = 2 s? (a) 16 m (b) 20 m (c) 24 m (d) 32 m (e) 72 m a ball of mass 2.0 kg is thrown vertically upward from the top of a building. the ball’s velocity v is given as a function of time t by the equation v(t)=r - st, where r = 5 m/s and s = 10 m/s^2. the positive direction is upward. 8. what is the ball’s acceleration at time t = 1 s? (a) -10 m/s^2 (b) -5 m/s^2 (c) 0 (d) 5 m/s^2 (e) 10 m/s^2 9. if the ball reaches the ground at t = 2.0 s, what is the height of the building? (a) 5 m (b) 10 m (c) 15 m (d) 20 m (e) 30 m 10. at what time does the ball reach its maximum height above the building? (a) 0.2 s (b) 0.4 s (c) 0.5 s (d) 1.0 s (e) 2.0 s

Explanation:

Step1: Recall the relationship between acceleration and velocity

Acceleration $a(t)$ is the derivative of velocity $v(t)$ with respect to time. For $v(t)=4 + 2t^{3}$, using the power - rule $\frac{d}{dt}(x^{n})=nx^{n - 1}$, we have $a(t)=\frac{dv}{dt}=6t^{2}$.

Step2: Calculate the acceleration at $t = 2s$

Substitute $t = 2$ into $a(t)=6t^{2}$. So $a(2)=6\times2^{2}=6\times4 = 24m/s^{2}$.

Step1: Recall the relationship between displacement and velocity

The displacement $x$ of an object moving with velocity $v(t)$ over the interval $[a,b]$ is given by $x=\int_{a}^{b}v(t)dt$. Here, $v(t)=4 + 2t^{3}$, $a = 0$, and $b = 2$.

Step2: Calculate the definite - integral

$\int_{0}^{2}(4 + 2t^{3})dt=\int_{0}^{2}4dt+\int_{0}^{2}2t^{3}dt$.
$\int_{0}^{2}4dt=4t\big|_{0}^{2}=4\times(2 - 0)=8$.
$\int_{0}^{2}2t^{3}dt=2\times\frac{t^{4}}{4}\big|_{0}^{2}=\frac{1}{2}t^{4}\big|_{0}^{2}=\frac{1}{2}(2^{4}-0^{4})=\frac{1}{2}\times16 = 8$.
Then $\int_{0}^{2}(4 + 2t^{3})dt=8 + 8=16m$.

Step1: Identify the acceleration from the velocity function

Given $v(t)=R - St$ where $R = 5m/s$ and $S = 10m/s^{2}$, acceleration $a(t)=\frac{dv}{dt}=-S$. So $a(t)=- 10m/s^{2}$ (constant acceleration), and at $t = 1s$, $a=-10m/s^{2}$.

Step1: Recall the displacement formula

The displacement $y - y_{0}=v_{0}t-\frac{1}{2}gt^{2}$ (in the vertical - motion context, here $v_{0}=R = 5m/s$, $g = S=10m/s^{2}$, and $y - y_{0}=-h$ where $h$ is the height of the building).
The displacement formula can also be obtained from integrating the velocity function $v(t)=R - St$. $\int_{0}^{t}v(t)dt=\int_{0}^{t}(R - St)dt=Rt-\frac{1}{2}St^{2}$.
When $t = 2s$, $\int_{0}^{2}(5-10t)dt=5t-5t^{2}\big|_{0}^{2}=5\times2-5\times2^{2}=10 - 20=-10m$. The height of the building $h = 10m$.

Answer:

D. $24m/s^{2}$