Math Snap

STEP 1

1. The mass of the hawk is 1 kg.
2. The initial speed of the hawk at time $t_1$ is $20 \, \mathrm{m/s}$.
3. The final speed of the hawk at time $t_2$ is $50 \, \mathrm{m/s}$.
4. We are asked to find the change in kinetic energy of the hawk between $t_1$ and $t_2$.

STEP 2

1. Recall the formula for kinetic energy.
2. Calculate the initial kinetic energy at $t_1$.
3. Calculate the final kinetic energy at $t_2$.
4. Determine the change in kinetic energy.

STEP 3

Recall the formula for kinetic energy, which is given by:
$$KE = \frac{1}{2} m v^2$$ where $m$ is the mass and $v$ is the velocity.

STEP 4

Calculate the initial kinetic energy at $t_1$ using the formula:
$$KE_1 = \frac{1}{2} \times 1 \, \mathrm{kg} \times (20 \, \mathrm{m/s})^2$$ $$KE_1 = \frac{1}{2} \times 1 \times 400$$ $$KE_1 = 200 \, \mathrm{J}$$

STEP 5

Calculate the final kinetic energy at $t_2$ using the formula:
$$KE_2 = \frac{1}{2} \times 1 \, \mathrm{kg} \times (50 \, \mathrm{m/s})^2$$ $$KE_2 = \frac{1}{2} \times 1 \times 2500$$ $$KE_2 = 1250 \, \mathrm{J}$$

Determine the change in kinetic energy by subtracting the initial kinetic energy from the final kinetic energy:
$$\Delta KE = KE_2 - KE_1$$ $$\Delta KE = 1250 \, \mathrm{J} - 200 \, \mathrm{J}$$ $$\Delta KE = 1050 \, \mathrm{J}$$ The change in kinetic energy of the hawk between $t_1$ and $t_2$ is most nearly:
$\boxed{1050 \, \mathrm{J}}$