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Get Full Access to University Physics - 13 Edition - Chapter 2 - Problem 45e
Get Full Access to University Physics - 13 Edition - Chapter 2 - Problem 45e

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# BIO The rocket-driven sled Sonic Wind No. 2, used for ISBN: 9780321675460 31

## Solution for problem 45E Chapter 2

University Physics | 13th Edition

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Problem 45E

BIO? The rocket-driven sled ?Sonic Wind No. 2,? used for investigating the physiological effects of large accelerations, runs on a straight, level track 1070 m (3500 ft) long. Starting from rest, it can reach a speed of 224 m/s (500 mi/h) in 0.900 s. (a) Compute the acceleration in m/s2, assuming that it is constant. (b) What is the ratio of this acceleration to that of a freely falling body (g)? (c) What distance is covered in 0.900 s? (d) A magazine article states that at the end of a certain run, the speed of the sled de-creased from 283 m/s (632 mi/h) to zero in 1.40 s and that during this time the magnitude of the acceleration was greater than 40 g . Are these figures consistent?

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Solution 45E Initial speed u = 0 Final speed v = 224 m/s Time t = 0.900 s From the equation v = u + at 224 = 0 + a × 0.900 2 a = 224/0.900 m/s 2 a = 249 m/s (a) Therefore, the acceleration is 249 m/s .2 The acceleration of a freely falling body is 9.8 m/s .2 2 2 (b) Therefore, the ratio of 249 m/s to 9.8 m/s is = 249/9.8 = 25.4 Time t = 0.900 s (c) Distance covered = at 2 2 Distance covered = 1 × 249 × (0.900) m2 2 Distance covered = 100.84 m (d) Initial speed = 283 m/s Final speed = 0 Time = 1.40 s Acceleration = Change of speed/ Time Acceleration = (0 283)/1.40 m/s 2 2 Acceleration = 202 m/s 2 Acceleration = 202/9.8 m/s Acceleration = 20.62 g This acceleration value is almost half of their given value of 40g. So, the figures are inconsistent.

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##### ISBN: 9780321675460

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