Question

(a) How long will it take an 850-kg car with a useful power output of 40.0 hp (1 hp equals 746 W) to reach a speed of 15.0 m/s, neglecting friction? (b) How long will this acceleration take if the car also climbs a 3.00-m high hill in the process?

Answer 1

Answer:

(A) time = 3.205 s

(B)time =4.04 s

Explanation:

mass (m) = 850 kg

power (P) = 40 hp = 40 x 746 = 29,840 W

final velocity (Vf) =  15 m/s

final height (Hf) = 3 m

since the car is starting from rest at the bottom of the hill, its initial velocity and initial height are both 0

(A) from the work energy theorem

• work = 0.5 x m x ($(Vf)^{2} - (Vi)^{2}$)  (change in kinetic energy)
• work = power x time
• therefore

        power x time = 0.5 x m x ($(Vf)^{2} - (Vi)^{2}$)

        time = $\frac{0.5 x m x ([tex](Vf)^{2} - (Vi)^{2}$)}{power}[/tex]

time = $\frac{0.5 x 850 x ([tex](15)^{2} - (0)^{2}$)}{29,840}[/tex]

time = 3.205 s

(B) from the work energy theorem

• work = change in potential energy + change in kinetic energy

• work = (mg (Hf - Hi)) + (0.5m($(Vf)^{2} - (Vi)^{2}$)

• work = power x time
• therefore

      power x time = (mg (Hf - Hi)) + (0.5m($(Vf)^{2} - (Vi)^{2}$)

      time = $\frac{(mg (Hf - Hi)) + (0.5m([tex](Vf)^{2} - (Vi)^{2}$)}[/tex])}{power}[/tex]

     

time = $\frac{(850 x 9.8 x (3 - 0)) + (0.5 x 850 x [tex](15)^{2} - (0)^{2}$)}[/tex])}{29,840}[/tex]

time =4.04 s

Answer 2

Answer:

a) $\Delta t = 3.205\,s$, b) $\Delta t = 4.043\,s$

Explanation:

a) The time needed is determined by the Work-Energy Theorem and the Principle of Energy Conservation:

$K_{1} + \Delta E = K_{2}$

$\Delta E = K_{2} - K_{1}$

$\dot W \cdot \Delta t = \frac{1}{2}\cdot m \cdot v^{2}$

$\Delta t = \frac{m\cdot v^{2}}{2\cdot \dot W}$

$\Delta t = \frac{(850\,kg)\cdot \left(15\,\frac{m}{s} \right)^{2}}{2\cdot (40\,hp)\cdot \left(\frac{746\,W}{1\,hp} \right)}$

$\Delta t = 3.205\,s$

b) The time is found by using the same approach of the previous point:

$U_{1} + K_{1} + \Delta E = U_{2} + K_{2}$

$\Delta E = (U_{2}-U_{1})+(K_{2} - K_{1})$

$\dot W \cdot \Delta t = m\cdot \left(g\cdot \Delta h + \frac{1}{2}\cdot v^{2} \right)$

$\Delta t = \frac{m\cdot\left(g\cdot \Delta h + \frac{1}{2}\cdot v^{2}\right)}{\dot W}$

$\Delta t = \frac{(850\,kg)\cdot \left[\left(9.807\,\frac{m}{s^{2}} \right)\cdot (3\,m) + \frac{1}{2}\cdot \left(15\,\frac{m}{s} \right)^{2}\right]}{(40\,hp)\cdot \left(\frac{746\,W}{1\,hp} \right)}$

$\Delta t = 4.043\,s$