HW1-phy2048-Spring-2014-solution

# 0 ms2 d thinking of xt in terms of accumulated area

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Unformatted text preview: + at ⇒ t = = 5.00 s . − 4.92 m/s 2 (b) Although several of the equations in Table 2- 1 will yield the result, we choose Eq. 2- 16 (since it does not depend on our answer to part (a)). (24.6 m/s) 2 2 0 = v0 + 2ax ⇒ x = − = 61.5 m . 2 − 4.92 m/s 2 ( ) 5 PHY2048 Spring 2014 HW1 2P38. We assume the train accelerates from rest ( v0 = 0 and x0 = 0 ) at a1 = +1.34 m / s2 until the midway point and then decelerates at a2 = −1.34 m / s2 until it comes to a stop ( v2 = 0 ) at the next station. The velocity at the midpoint is v1, which occurs at x1 = 806/2 = 403m. (a) Equation 2- 16 leads to 2 v12 = v0 + 2a1 x1 ⇒ v1 = 2 (1.34 m/s 2 ) ( 403 m ) = 32.9 m/s. (b) The time t1 for the accelerating stage is (using Eq. 2- 15) x1 = v0t1 + 2 ( 403 m ) 12 a1t1 ⇒ t1 = = 24.53 s . 2...
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