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A car is travelling at $ 50\;mi/h $ when the brakes are fully applied, producing a constant deceleration of $ 22\;ft/s^2 $. What is the distance traveled before the car comes to a stop?

$$

\approx 122.22

$$

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we know that we have the prime of cheese of the derivative of velocity, which we know is equivalent to accelerations. Empty is negative 22 therefore, V A t. The integral of acceleration is negative. 22 times are variable of tea plus c. Therefore, if we stop via oh, is CR constant, you know see is 220 divided by three which gives our equation V of tea is equivalent to negative 22 tea plus 2 20 divided by three us of tea is negative. 11 t squared was 220 over three. Chief trustee and then we have zero is negative. 22 tea lost 220 over three and then t is 10 over three, which gives us night of 11 times 10 over three square was 220 over three times 10 over three, which some flies to 122.22 feet. That's our section