QuestionAn automobile accelerated from rest to $345.6 \mathrm{~m/s}$ in $3.07 \mathrm{~s}$ . What distance did it travel?

Studdy Solution

STEP 1

Assumptions1. The car starts from rest, so its initial velocity is $0 \mathrm{~m/s}$ . . The final velocity of the car is $345.6 \mathrm{~m/s}$ .
3. The time taken to reach this speed is $3.07 \mathrm{~s}$ .
4. The car travels a certain distance in this time, which we need to find.
5. The acceleration is constant.

STEP 2

We can use the equation of motion to find the distance traveled by the car. The equation is $s = ut + \frac{1}{2}at^2$ where- $s$ is the distance traveled, - $u$ is the initial velocity, - $a$ is the acceleration, and- $t$ is the time.

STEP 3

Since the car starts from rest, the initial velocity $u$ is $0 \mathrm{~m/s}$ . So the equation simplifies to $s = \frac{1}{2}at^2$

STEP 4

We need to find the acceleration first. The formula for acceleration is $a = \frac{v - u}{t}$ where- $v$ is the final velocity, - $u$ is the initial velocity, and- $t$ is the time.

STEP 5

Substitute the given values into the acceleration formula $a = \frac{345. \mathrm{~m/s} -0 \mathrm{~m/s}}{3.07 \mathrm{~s}}$

STEP 6

Calculate the acceleration $a = \frac{345.6 \mathrm{~m/s}}{3.07 \mathrm{~s}} =112.6 \mathrm{~m/s^2}$

STEP 7

Now that we have the acceleration, we can substitute it into the equation for distance $s = \frac{1}{2} \times112.6 \mathrm{~m/s^2} \times (3.07 \mathrm{~s})^2$

STEP 8

Calculate the distance $s = \frac{1}{2} \times112.6 \mathrm{~m/s^2} \times.424 \mathrm{~s^2} =530.5 \mathrm{~m}$ The car traveled530.5 meters.

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