Saturday

November 29, 2014

November 29, 2014

Posted by **Kyle** on Monday, June 16, 2014 at 10:07pm.

interval of 4.08 seconds.

(a.) What is the average acceleration of the car during this time?

(b.) Assuming the acceleration is constant, how far did the car travel during this time?

- General Physics 1 -
**Damon**, Monday, June 16, 2014 at 10:23pmchange in velocity/ change in time = (17.9-29.6) / 4.08

= -2.67 m/s^2

d = Vi t - (1/2) 2.67 * t^2

= 29.6(4.08) - 1.33 (4.08^2)

= 121 - 22.1

= 98.9 m

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