Question

thomson’s gazelles are some of the fastest land animals on earth. it’s estimated they can run with a speed of $24\frac{\text{m}}{\text{s}}$ (about $54\frac{\text{mi}}{\text{h}}$). gazelles do not accelerate as quickly as some of their predators, but they can maintain their speed for a longer time. if a gazelle begins at rest and begins running with an acceleration of $4\frac{\text{m}}{\text{s}^2}$, how long will it take him to reach a speed of $24\frac{\text{m}}{\text{s}}$?

Explanation:

Step1: Recall the kinematic equation

We know that the kinematic equation relating final velocity \( v \), initial velocity \( u \), acceleration \( a \), and time \( t \) is \( v = u + at \).
Here, the gazelle starts from rest, so the initial velocity \( u = 0\space\frac{\text{m}}{\text{s}} \), the final velocity \( v = 24\space\frac{\text{m}}{\text{s}} \), and the acceleration \( a = 4\space\frac{\text{m}}{\text{s}^2} \).

Step2: Rearrange the equation to solve for time

From \( v = u + at \), since \( u = 0 \), the equation simplifies to \( v = at \). We can rearrange this to solve for \( t \): \( t=\frac{v}{a} \).

Step3: Substitute the values

Substitute \( v = 24\space\frac{\text{m}}{\text{s}} \) and \( a = 4\space\frac{\text{m}}{\text{s}^2} \) into the formula for \( t \):
\( t=\frac{24\space\frac{\text{m}}{\text{s}}}{4\space\frac{\text{m}}{\text{s}^2}} \)
When we divide the numerator and the denominator, the units of meters cancel out, and we have \( \frac{\text{s}^2}{\text{s}}=\text{s} \) for the units. Calculating the numerical value: \( \frac{24}{4} = 6 \).

Answer:

The time taken for the gazelle to reach a speed of \( 24\space\frac{\text{m}}{\text{s}} \) is \( \boldsymbol{6\space\text{seconds}} \).