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An automobile initially moving at 30 ft/sec accelerates uniformly at 15 ft/sec^2. How...

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else9393 | Student, College Freshman | eNoter

Posted October 26, 2010 at 2:43 AM via web

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An automobile initially moving at 30 ft/sec accelerates uniformly at 15 ft/sec^2. How fast is it moving after 3 sec.

At the end of 3s the driver hits the brakes and the automobile now accelerates at -30 ft/sec^2. How long does it take to come to a complete stop?

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william1941 | College Teacher | Valedictorian

Posted October 26, 2010 at 2:45 AM (Answer #1)

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We are given that the initial velocity of the automobile is 30 ft/s. The rate at which it is accelerating is 15 ft/s^2.

The relation between the initial velocity V0 and final velocity Vt after an accelaration of 'a' for 't' sec is given as Vt = V0 + at

Therefore the speed after 3s is given by 30 + 15*3 = 30 + 45 = 75 ft/s.

After 3 seconds the automobile starts to decelerate as the brakes are applied. The rate of deceleration is 30 ft/s^2.

Therefore as it was going at 75 ft/s, the time taken to come to a stop is (75 - 0) / 30 = 2.5 s.

The automobile takes 2.5 s to stop completely.

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giorgiana1976 | College Teacher | Valedictorian

Posted October 26, 2010 at 2:44 AM (Answer #2)

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We'll have to write the equation that describes the velocity of an object under acceleration.

v = v0 + a*t (1)

We'll insert the given informations:

v(3s) = 30 ft/s + (15 ft/s^2)(3s)

v(3s) = 30 ft/s + 45 ft/s

v(3s) = 75 ft/s

At the end of the interval of 3s, the car is decelerated (the acceleration is opposite to the direction of velocity).

v = 75 ft/s + (-30 ft/s^2)t

When the car comes to stop, v = 0.

0 = 75 ft/s + (-30 ft/s^2)t

-75 ft/s = -30(ft/s^2)*t

t = 2.5 s

After 2.5 s, the car comes to stop.

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