Given speed of pickup truck, v₁ = 90 km/h.

Given speed of pickup truck, $v_1 = 90 \text{ km/h}$.

$$v_1 = 90 \text{ km/h} \times \frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}}$$ $$v_1 = \frac{90 \times 1000}{3600} \text{ m/s}$$ $$v_1 = 25 \text{ m/s}$$

Given mass of pickup truck, $m_1 = 2000 \text{ kg}$.
The formula for kinetic energy is $KE = \frac{1}{2}mv^2$.

$$KE_1 = \frac{1}{2}m_1 v_1^2$$ $$KE_1 = \frac{1}{2}(2000 \text{ kg})(25 \text{ m/s})^2$$ $$KE_1 = \frac{1}{2}(2000 \text{ kg})(625 \text{ m}^2/\text{s}^2)$$ $$KE_1 = 1000 \times 625 \text{ J}$$ $$KE_1 = 625000 \text{ J}$$

Given mass of sport car, $m_2 = 1000 \text{ kg}$.
We are given that $KE_2 = KE_1$.

$$KE_2 = 625000 \text{ J}$$

Using the kinetic energy formula for the sport car:

$$KE_2 = \frac{1}{2}m_2 v_2^2$$ $$625000 \text{ J} = \frac{1}{2}(1000 \text{ kg})v_2^2$$ $$625000 = 500 v_2^2$$ $$v_2^2 = \frac{625000}{500}$$ $$v_2^2 = 1250$$ $$v_2 = \sqrt{1250} \text{ m/s}$$ $$v_2 \approx 35.355 \text{ m/s}$$

$$v_2 = 35.355 \text{ m/s} \times \frac{3600 \text{ s}}{1 \text{ h}} \times \frac{1 \text{ km}}{1000 \text{ m}}$$ $$v_2 = \frac{35.355 \times 3600}{1000} \text{ km/h}$$ $$v_2 \approx 127.278 \text{ km/h}$$

The speed of the sport car is approximately $35.4 \text{ m/s}$ or $127.3 \text{ km/h}$.