Question

When a 1984 Alfa Romeo Spider sports car accelerates at the maximum possible rate, its motion during the first 20 s is extremely well modeled by the simple equation
${v_x}^2=\frac{2P}{m}t$
where P = 3.6 * 10⁴ watts is the car’s power output, m = 1200 kg is its mass and$v_x$ is in m/s. That is, the square of the car’s velocity increases linearly with time.
a. Find an algebraic expression in terms of P, m, and t for the car’s acceleration at time t.
b. What is the car’s speed at t = 2 s and t = 10 s?
c. Evaluate the acceleration at t = 2 s and t = 10 s

Short answer

a) The acceleration of the car is $\sqrt{\frac{P}{2mt}}$
b) The velocity at time t = 2 sec and 10 sec is 10.9 m/s and 24.5 m/s respectively
c) The acceleration of the car at time t= 2 sec and 10 sec is 2.7 m/s² and 1.22 m/s² respectively.

Step-by-step solution

Step 1. Write the given information

The power of the car is $P=3.6\times {10}^{4}watt$
The mass of the car is$m=2400kg$

The expression of velocity is given as

role="math" localid="1648619581365" $$\begin{gathered} {v_x}^2=\frac{2P}{m}t \\ {v}_x=\sqrt{\frac{2P}{m}t}.....\left(1\right) \end{gathered}$$

Step 2. (a) To determine the acceleration of the car

The acceleration of the body is the change in velocity with respect to time. thus, the acceleration of the car is given by
$$\begin{gathered} \frac{d}{dt}\left({v_x}^2\right)=\frac{d}{dt}\left[\frac{2P}{m}t\right] \\ 2v_x\left(\frac{d{v}_x}{dt}\right)=\frac{2P}{m} \\ 2v_{x}a=\frac{2P}{m} \\ a=\frac{P}{m\left(v_x\right)} \end{gathered}$$
Plug the value of$v_x$ from equation (1) in the above equation
role="math" localid="1648619721540" $$\begin{gathered} a=\frac{P}{m\sqrt{{\displaystyle \frac{2P}{m}}t}} \\ a=\sqrt{\frac{P}{2mt}}......\left(2\right) \end{gathered}$$

Thus, the acceleration of the car in terms of P, m, and t is $\sqrt{\frac{P}{2mt}}$

Step 3. (b) To determine the velocity of the car

At time $t=2sec$, the velocity of the car is given by the equation (1)
$$\begin{gathered} v_x=\sqrt{\frac{2(3.6\times {10}^4)}{\left(1200\right)}\left(2\right)}=\sqrt{120} \\ {v}_x=10.9m/s \end{gathered}$$

Similarly, at time $t=10sec$, the velocity of the car is given by

$$\begin{gathered} v_x=\sqrt{\frac{2(3.6\times {10}^4)}{1200}\left(10\right)}=\sqrt{600} \\ {v}_x=24.5m/s \end{gathered}$$

Step 4. (c) To determine the acceleration of the car

At time$t=2sec$, the acceleration of the car is given by equation (2)
$$\begin{gathered} a=\sqrt{\frac{3.6\times {10}^4}{2\left(1200\right)\left(2\right)}}=\sqrt{7.5} \\ a=2.7m/s^2 \end{gathered}$$

Similarly, at time$t=10sec$, the acceleration of the car
$$\begin{gathered} a=\sqrt{\frac{3.6\times {10}^4}{2\left(1200\right)\left(10\right)}}=\sqrt{1.5} \\ a=1.22m/s^2 \end{gathered}$$