Let's break down the problem step by step.

Given Information:

• Mass of the car ($m$): 850 kg
• Power ($P$): 40 HP (convert to watts, where $1 \text{ HP} = 745.7 \text{ W}$, so $40 \text{ HP} = 40 \times 745.7 \, \text{W}$).
• Final velocity ($v$): 15 m/s
• Neglect friction for part (a).
• Height of the hill ($h$): 3 m (for part b).

(a) How long does it take to reach a velocity of 15 m/s?

We can use the definition of power to link it with kinetic energy.

1. Kinetic energy of the car when it reaches 15 m/s: $E_k = \frac{1}{2} m v^2$ Substituting values: $E_k = \frac{1}{2} \times 850 \times (15)^2 = 95625 \, \text{J}$

2. Power available: Convert 40 HP to watts: $P = 40 \times 745.7 = 29828 \, \text{W}$

3. Time to reach this velocity: Power is the rate of doing work (or energy per time), so: $P = \frac{E_k}{t} \quad \Rightarrow \quad t = \frac{E_k}{P} = \frac{95625}{29828} \approx 3.21 \, \text{seconds}$

(b) How long does it take if the car also climbs a 3-meter hill simultaneously?

In this case, we need to account for both the kinetic energy and the potential energy the car gains by climbing the hill.

1. Potential energy gained by the car when it climbs the hill: $E_p = mgh$ Substituting values: $E_p = 850 \times 9.8 \times 3 = 24990 \, \text{J}$

2. Total energy needed (kinetic + potential): $E_{total} = E_k + E_p = 95625 + 24990 = 120615 \, \text{J}$

3. Time to reach 15 m/s while climbing: $t = \frac{E_{total}}{P} = \frac{120615}{29828} \approx 4.04 \, \text{seconds}$

Thus, the time for part (a) is approximately 3.21 seconds, and for part (b), it is 4.04 seconds, matching the provided answers.

Would you like more details or have any questions?

Follow-up Questions:

1. How would friction affect the time to reach the final velocity?
2. What would happen if the car’s mass were doubled?
3. How would the time change if the hill were twice as high?
4. What if the power of the car was reduced by half?
5. How does the result change if the car starts from an initial velocity?

Tip: Always ensure to convert units consistently, especially when dealing with power (e.g., HP to watts).