Question

On a dry road a car with good tires may be able to brake with a deceleration of $11.0\mathrm{mi}/\mathrm{h}/\mathrm{s}(=4.92\mathrm{m}/{\mathrm{s}}^2).(a)$ How long does it take such a car, initially traveling at $55\mathrm{mi}/\mathrm{h}(=24.6$ $\mathrm{m}/\mathrm{s}$ ), to come to rest? $(b)$ How far does it travel in this time?

Short answer

The time the car takes to stop is approximately $5$ seconds and the distance travelled in this time is about $61.5$ meters.

Step-by-step solution

Identify given information

The initial speed of the car is given as $24.6m/s$. The deceleration or negative acceleration is given as $-4.92m/s^2$ (it is negative since it's slowing down the car). The final speed of the car would be $0m/s$ as it comes to rest.

Calculate time taken to stop

We use the first equation of motion, which is $v=u+at$, where $v$ is the final speed, $u$ is the initial speed, $a$ is the acceleration and $t$ is the time taken. Rearranging for $t$, we get $t=(v-u)/a$. Substituting the given values, we calculate $t=(0-24.6)/-4.92$.

Compute the distance traveled

For this part, we use the third equation of motion, which is $v^2=u^2+2as$, where $s$ is the distance traveled. Rearranging for $s$, we get $s=(v^2-u^2)/(2a)$. Substituting the values, we compute $s=[(0{)}^2-(24.6{)}^2]/[2\ast (-4.92)]$.

Key concepts

Equations of Motion

Equations of motion are at the heart of understanding how objects move and stop. They are mathematical relationships that describe the behavior of a moving object under certain conditions like speed, acceleration, and time. Here's a look at two key equations:

• The First Equation of Motion:

  This is expressed as $v=u+at$. In this equation:

  • $v$ is the final velocity of an object,
  • $u$ is the initial velocity,
  • $a$ is the acceleration, which can be positive or negative,
  • $t$ is the time it takes to change from initial to final velocity.
  By rearranging it, as we see in the step-by-step solution, we can find out how long it takes for a car to stop given its starting speed and deceleration.
• The Third Equation of Motion:

  This one is given by $v^2=u^2+2as$. It relates final and initial velocities, acceleration, and the distance $s$ traveled. Using this, it becomes possible to discover how far the car travels before coming to a stop.

These equations help to form a comprehensive picture of motion in physics, especially when dealing with constant acceleration, such as deceleration in braking scenarios.

Initial and Final Velocity

Understanding initial and final velocity is critical in solving motion problems. Initial velocity $u$ is simply the speed at which an object, like a car, starts moving in a given situation. In this exercise, the car starts at an initial velocity of 24.6 m/s.

Final velocity $v$ is the speed of the object after a period of acceleration (or deceleration). Here, the car comes to a stop, which means the final velocity is 0 m/s.

The change from initial to final velocity is governed by the acceleration (negative in the case of deceleration). By using equations of motion, you can calculate other unknowns, such as time taken to stop or distance traveled.

Distance Calculation

Calculating the distance a car travels while braking involves using equations of motion. In the exercise, the initial speed is 24.6 m/s, and the final speed is 0 m/s, which simplifies the understanding of the movement.

To find this distance, we use the third equation of motion: $s=\frac{v^2-u^2}{2a}$. Here, $v^2$ and $u^2$ are the squares of the final and initial velocities, respectively. The acceleration $a$, which is negative in deceleration, plays a crucial role in this calculation.

• This equation allows solving for $s$, the distance, in a straightforward manner by plugging in known values. In our example, this calculation gives us the precise distance the car travels during braking.
• Understanding this is vital for real-world applications, like ensuring safe stopping distances for vehicles.