Problems 69 t two frames 63. Il Scientists have studied two (m/s he track has sp

Question

Problems 69 t two frames 63. Il Scientists have studied two (m/s he track has species of sand lizards, the 2.0 Mojave fringe-toed lizard and 1 1.0 the western zebra-tailed liz- os enter of the 1, showing ccurred. you have ould con- 0.5 ard, to understand the extent 00 s0 00 15o 20% 25 Cms) to which the different struc- ture of the two species' toes FIGURE P2.63 is related to their preferred habitats-fine sand for the Mojave lizard and coarse sand for the zebra-tailed lizard. Figure P2.63 shows a somewhat simplified velocity-versus-time graph for the Mojave fringe-toed lizard. 2.0 s to a. Estimate the maximum acceleration of the lizard in both m/s and g's section? b. Estimate its acceleration at t 150 ms. ction? n of the 64. I You are driving to the grocery store at 20 m/s. You are 110 c. Estimate how far it travels in the first 50 ms. m from an intersection when the traffic light turns red. Assume that your reaction time is 0.70 s and that your car brakes with constant acceleration. a. How far are you from the intersection when you begin to apply the brakes? b. What acceleration will bring you to rest right at the intersection? c. How long does it take you to stop? 65. I When you blink your eye, the upper lid goes from rest with your BIO eye open to completely covering your eye in a time of 0.024 s. a. Estimate the distance that the top lid of your eye moves during a blink. b. What is the acceleration of your eyelid? Assume it to be constant. c. What is your upper eyelid's final speed as it hits the bottom 0 m. min eyelid? she 66. ll A bush baby, an African primate, is an- BIO capable of a remarkable vertical leap. The huch hahy goes into a crouch and extends

Explanation / Answer

a) as my reaction time is 0.7 s

during this time i will move distance d = 0.7 x 20 = 14 m

so now when i will apply breaks i will be ( 110-14) m = 96 m from intersection point

b) now to stop in this 96 m distance lets say retardation is a m/s2

u = 20 m/s

s = 96 m

from newtons 3rd eq

v2 =u2 + 2as

0 = 202 +2xax96

a = - 2.08 m/s2

c) lets assume time taken to stop is = t

from

v = u +at

0 = 20 - 2.08 t

t = 9.62 sec

answer

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