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An automobile initially moving at 30 ft/sec accelerates uniformly at 15 ft/sec^2. How...
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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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.
Posted by william1941 on October 26, 2010 at 2:45 AM (Answer #1)
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.
Posted by giorgiana1976 on October 26, 2010 at 2:44 AM (Answer #2)
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