High School

A driver in a car, originally moving at 11.8 m/s, applies the brakes until the car comes to a stop. The car covers a distance of 34.5 m while braking. How much time did it take for the car to stop? Assume constant acceleration during braking.

Answer :

The time it took for the car to stop is equal to the acceleration.

To determine the time it took for the car to stop, we can use the formula:

[tex]\[v^2 = u^2 + 2as\][/tex]

Since we want to solve for time, we can rearrange the formula to:
[tex]\[t = \frac{v-u}{a}\][/tex]

Since [tex]\(v = 0\)[/tex], the formula simplifies to:
[tex]\[t = -\frac{u}{a}\][/tex]

Now we can substitute the given values into the formula. The initial velocity u is 11.8 m/s and the distance traveled s is 34.5 m. However, we still need to find the acceleration a.

To find the acceleration, we can use the formula:
[tex]\[a = \frac{v-u}{t}\][/tex]

Since we know the final velocity v is 0 m/s, the formula simplifies to:
[tex]\[a = -\frac{u}{t}\][/tex]

Now we can rearrange the formula to solve for acceleration:
[tex]\[a = -\frac{u}{t}\][/tex]

Substituting the given values, we have:
[tex]\[a = -\frac{11.8}{t}\][/tex]

Now we can substitute this value of a back into the equation for time:
[tex]\[t = -\frac{u}{a} = -\frac{11.8}{-\frac{11.8}{t}} = t\]\\[/tex]

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