Question

If a 1300 kg car can accelerate from 30 km/h to 65 km/h in 3.6 s , how long will it take to accelerate from 65 km/h to 90 km/h ? Assume the power stays the same, and neglect frictional losses.
Express your answer to two significant figures and include the appropriate units.

Answer 1

Let's first convert km/h to m/s:

$$v_f = 65 \dfrac{km}{h} \cdot \dfrac{1000m}{1 km} \cdot \dfrac{1 h}{3600 s} = 18.05 \dfrac{m}{s},$$

$$v_i = 30 \dfrac{km}{h} \cdot \dfrac{1000m}{1 km} \cdot \dfrac{1 h}{3600 s} = 18.05 \dfrac{m}{s}.$$

By the definition of the power, we get:

$$P = \dfrac{W}{t},$$

here, $W$ is the work done by the car and $t$ is time.

Let's find the work done by the car from the Work-Kinetic Energy theorem. It states that the work done by the car is equal to the change in kinetic energy of the car:

$$W = \Delta KE = KE_f - KE_i,$$ $$W = \dfrac{1}{2}mv_f^2 - \dfrac{1}{2}mv_i^2 = \dfrac{1}{2}m(v_f^2 - v_i^2).$$

Substituting $W$ into the first formula we can find the power of the car when it accelerates from 30 km/h to 65 km/h in 3.6 s:

$$P = \dfrac{\dfrac{1}{2} \cdot 1300kg \cdot((18.05 \dfrac{m}{s})^2 - (8.3 \dfrac{m}{s})^2)}{3.6 s} = 46387 W.$$

Now, let's find the work done by the car when it accelerates from 65 km/h to 90 km/h from the same Work-Kinetic Energy theorem, but, first, convert km/h to m/s:

$$v_f = 90 \dfrac{km}{h} \cdot \dfrac{1000m}{1 km} \cdot \dfrac{1 h}{3600 s} = 25 \dfrac{m}{s},$$ $$v_i = 65 \dfrac{km}{h} \cdot \dfrac{1000m}{1 km} \cdot \dfrac{1 h}{3600 s} = 18.05 \dfrac{m}{s},$$ $$W = \dfrac{1}{2} \cdot 1300 kg \cdot ((25 \dfrac{m}{s})^2 - (18.05 \dfrac{m}{s})^2) = 194478 J.$$

Since, the power stays the same we can find the time that the car takes to accelerate from 65 km/h to 90 km/h from the same formula for the power:

$$P = \dfrac{W}{t},$$ $$t = \dfrac{W}{P} = \dfrac{194478 J}{46387 W} = 4.2 s.$$

Answer:

$t = 4.2 s.$