Question

a car needs to stop before it flies off a cliff, if it is traveling at a velocity of 100 mph before slamming on its brakes and it breaks 75 feet away from the edge. will the car be able to stop in time?

Answer 1

The calculated braking distance is approximately
`\(360.15 \, \text{ft}\)`. Since this value is greater than the given braking distance of
`\(75 \, \text{ft}\)`, the car will not stop in time to avoid going off the cliff under these assumptions.






To determine whether the car will be able to stop in time to avoid going off the cliff, we need to consider the physics of motion and braking distance. The key concept here is the relationship between velocity, acceleration, and distance covered during braking.

The relevant physics equation for motion with constant acceleration is:

`\[ v^2 = u^2 + 2as \]`

where:

- \( v \) is the final velocity (in this case, the car comes to a stop, so \( v = 0 \)),

- \( u \) is the initial velocity,- \( a \) is the acceleration (in this case, the deceleration due to braking), and

- \( s \) is the distance traveled during braking.

The goal is to find out whether the car can stop before reaching the cliff, so we're interested in the distance \( s \).

First, let's convert the initial velocity from mph to feet per second (1 mph = 1.47 ft/s):

`\[ u = 100 \, \text{mph} * 1.47 \, \text{ft/s per mph} \]`

Now, we can rearrange the motion equation to solve for \( s \):

`\[ s = \frac{{v^2 - u^2}}{{2a}} \]`

Substitute the known values:

`\[ s = \frac{{0 - (100 * 1.47)^2}}{{2a}} \]`

We also need to consider the braking distance given as 75 feet. If the car stops within this distance, it will avoid going off the cliff.

If the calculated braking distance (\( s \)) is less than or equal to the given braking distance (75 feet), the car will stop in time to avoid going off the cliff. If it's greater, the car will not stop in time.

let's perform the calculation.

First, we convert the initial velocity from mph to ft/s:

`\[ u = 100 \, \text{mph} * 1.47 \, \text{ft/s per mph} \]\[ u = 147 \, \text{ft/s} \]`

Now, we can use the motion equation:

`\[ s = \frac{{v^2 - u^2}}{{2a}} \]`

Since the car comes to a stop
`(\(v = 0\))`, the equation simplifies to:

`\[ s = \frac{{-u^2}}{{2a}} \]`

We also have the given braking distance
`(\(s_{\text{given}}\))` as 75 feet.

If \(s\) is less than or equal to
`\(s_{\text{given}}\)`, the car stops in time.

Let's calculate:

`\[ s = \frac{{-147^2}}{{2a}} \]`

Now, the value of \(a\) (acceleration) depends on the braking system of the car. For simplicity, let's assume a typical maximum deceleration for a car, which is around
`\(30 \, \text{ft/s}^2\)` during heavy braking.

`\[ s = \frac{{-147^2}}{{2 * 30}} \]\[ s \approx \frac{{-21609}}{{60}} \]\[ s \approx -360.15 \, \text{ft} \]`

The negative sign indicates the direction of motion, but in this context, we are only concerned with the magnitude.

The calculated braking distance is approximately
`\(360.15 \, \text{ft}\)`. Since this value is greater than the given braking distance of
`\(75 \, \text{ft}\)`, the car will not stop in time to avoid going off the cliff under these assumptions.

Keep in mind that the actual deceleration could vary, and this calculation provides a simplified analysis. If you have more specific information about the braking system or deceleration, you can substitute that value into the calculation for a more accurate result.

The probable question can be: A car is approaching the edge of a cliff at a velocity of 100 mph. The driver applies the brakes, and the car comes to a stop 75 feet away from the cliff's edge. Considering the braking distance and the initial velocity, analyze whether the car will be able to stop in time to avoid going off the cliff. Include relevant physics concepts and calculations to support your answer."