HW1-phy2048-Spring-2014-solution

1 s for the total time between start ups thus eq 2 2

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Unformatted text preview: 048 Spring 2014 HW1 (b) Plugging in t = 1 s, we obtain v(1 s) = (6.0 m/s2 )(1.0 s) − (6.0 m/s3 )(1.0 s)2 = 0. (c) Similarly, v(2 s) = (6.0 m/s 2 )(2.0 s) − (6.0 m/s3 )(2.0 s) 2 = −12m/s . (d) v(3 s) = (6.0 m/s 2 )(3.0 s) − (6.0 m/s3 )(3.0 s) 2 = − 36 m/s . (e) v(4 s) = (6.0 m/s2 )(4.0 s) − (6.0 m/s3 )(4.0 s) 2 = −72 m/s . The acceleration is given by a = dv/dt = 2c – 6b = 6.0 m/s2 – (12.0 m/s3)t. (f) Plugging in t = 1 s, we obtain a(1 s) = 6.0 m/s2 − (12.0 m/s3 )(1.0 s) = − 6.0 m/s2 . (g) a(2 s) = 6.0 m/s 2 − (12.0 m/s3 )(2.0 s) = − 18 m/s 2 . (h) a(3 s) = 6.0 m/s 2 − (12.0 m/s3 )(3.0 s) = −30 m/s 2 . (i) a(4 s) = 6.0 m/s 2 − (12.0 m/s3 )(4.0 s) = − 42 m/s 2 . 2P28. We take +x in the direction of motion, so v0 = +24.6 m/s and a = – 4.92 m/s2. We also take x0 = 0. (a) The time to come to a halt is found using Eq. 2- 11: 24.6 m/s 0 = v0...
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This document was uploaded on 03/24/2014.

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