QuestionA cheetah runs at $30 \mathrm{~m/s}$ , accelerates at $5 \mathrm{~m/s}^2$ for 7 seconds. Find the distance covered.

Studdy Solution

STEP 1

Assumptions1. The initial speed of the cheetah is $30 \mathrm{~m/s}$ . The acceleration of the cheetah is $5 \mathrm{~m/s}^{}$
3. The time during which the cheetah is accelerating is7 seconds4. We are asked to find the distance covered by the cheetah during this time

STEP 2

We can use the formula for the distance covered by an object under constant acceleration, which is given by $d = v_{i}t + \frac{1}{2}at^{2}$ where $d$ is the distance, $v_{i}$ is the initial velocity, $a$ is the acceleration, and $t$ is the time.

STEP 3

Now, plug in the given values for the initial velocity, acceleration, and time into the formula.
$d =30 \mathrm{~m/s} \times7 \mathrm{~s} + \frac{1}{2} \times5 \mathrm{~m/s}^{2} \times (7 \mathrm{~s})^{2}$

STEP 4

First, calculate the product of the initial velocity and time.
$30 \mathrm{~m/s} \times7 \mathrm{~s} =210 \mathrm{~m}$

STEP 5

Next, calculate the product of the acceleration and the square of the time, and then divide by2.
$\frac{1}{2} \times5 \mathrm{~m/s}^{2} \times (7 \mathrm{~s})^{2} = \frac{1}{2} \times5 \mathrm{~m/s}^{2} \times49 \mathrm{~s^{2}} =122.5 \mathrm{~m}$

STEP 6

Finally, add the two results together to find the total distance covered by the cheetah.
$d =210 \mathrm{~m} +122.5 \mathrm{~m}$

STEP 7

Calculate the total distance covered by the cheetah.
$d =210 \mathrm{~m} +122.5 \mathrm{~m} =332.5 \mathrm{~m}$ The cheetah covers a distance of332.5 meters while chasing its prey.

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