Question

• written response (29 marks total)

he following questions in the spaces provided. for full credit, you must include formula and it manipulations
orrect units. for science 10 purposes, answers should include 2 decimal places
. a 1600-kg vehicle moves with a velocity of 19.5 m/s. calculate the power required to reduce the velocity to
3.20 m/s in 11.0 s. (4 marks)

nce 10 mini

Explanation:

Step1: Recall the formula for kinetic energy

The kinetic energy of an object is given by $KE = \frac{1}{2}mv^2$, where $m$ is the mass and $v$ is the velocity. We need to find the change in kinetic energy, $\Delta KE$, which is the initial kinetic energy minus the final kinetic energy.
Initial kinetic energy: $KE_i = \frac{1}{2}m v_i^2$
Final kinetic energy: $KE_f = \frac{1}{2}m v_f^2$
So, $\Delta KE = KE_i - KE_f = \frac{1}{2}m(v_i^2 - v_f^2)$

Step2: Substitute the values

Given: $m = 1600\ kg$, $v_i = 19.5\ m/s$, $v_f = 3.20\ m/s$
$\Delta KE = \frac{1}{2} \times 1600 \times (19.5^2 - 3.20^2)$
First, calculate $19.5^2 = 380.25$ and $3.20^2 = 10.24$
Then, $19.5^2 - 3.20^2 = 380.25 - 10.24 = 370.01$
$\Delta KE = \frac{1}{2} \times 1600 \times 370.01 = 800 \times 370.01 = 296008\ J$ (since kinetic energy is in joules)

Step3: Recall the formula for power

Power $P$ is the rate of doing work (or the rate of change of energy), so $P = \frac{\Delta KE}{t}$, where $t$ is the time taken.
Given $t = 11.0\ s$
$P = \frac{296008}{11.0} \approx 26909.82\ W$ (or $26.91\ kW$ when rounded to two decimal places)

Answer:

The power required is approximately $\boxed{26909.82\ W}$ (or $26.91\ kW$)