Scientists have studied two species of sand lizards, the Mojave fringe-toed liza

Question

Scientists have studied two species of sand lizards, the Mojave fringe-toed lizard and the western zebra-tailed lizard, to understand the extent to which the different structure of the two species' toes is related to their preferred habitats-fine sand for the Mojave lizard and coarse sand for the zebra-tailed lizard. (Figure 1) shows a somewhat simplified velocity-versus-time graph for the Mojave fringe-toed lizard.

Part A

Estimate the maximum acceleration of the lizard in both m/s2 and g's.

Express your answers using two significant figures separated by commas.

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Part B

Estimate its acceleration at t = 150 ms.

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Part C

Estimate how far it travels in the first 50 ms.

Express your answer to two significant figures and include the appropriate units

t

1 of 1

Scientists have studied two species of sand lizards, the Mojave fringe-toed lizard and the western zebra-tailed lizard, to understand the extent to which the different structure of the two species' toes is related to their preferred habitats-fine sand for the Mojave lizard and coarse sand for the zebra-tailed lizard. (Figure 1) shows a somewhat simplified velocity-versus-time graph for the Mojave fringe-toed lizard.

Part A

Estimate the maximum acceleration of the lizard in both m/s2 and g's.

Express your answers using two significant figures separated by commas.

amax =

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Incorrect; Try Again; 6 attempts remaining; no points deducted

Your answer does not have the correct number of comma-separated terms.

Part B

Estimate its acceleration at t = 150 ms.

Estimate its acceleration at  = 150 . 5 m/s2 or 0.5g 1 m/s2 or 0.1g 10 m/s2 or g 2 m/s2 or 0.2g

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Incorrect; Try Again; 5 attempts remaining

Part C

Estimate how far it travels in the first 50 ms.

Express your answer to two significant figures and include the appropriate units

x =

t

Figure

1 of 1

V (m/s) 2.0 1.5 1.0 0.5 0.0 t (ms) 0 50 100 150 200 250

Explanation / Answer

part A:

maximum acceleration occurs at the initial portion of the graph

acceleration=slope of velocity vs time curve

=(0.75-0)/(50-0) (m/s)/ms

=15 m/s^2

=1.53*g

part B:

consider a small segment from t=100 ms to t=150 ms

acceleration=slope=(1.6-1.3)/(150-100) (m/s)/ms

=6 m/s^2

nearby answer is 5 m/s^2 or 0.5g

hence first option is correct.

part C:

acceleration in first 50 ms=15 m/s^2

then distance travelled=initial speed*time+0.5*acceleration*time^2

=0*0.050+0.5*15*0.05^2

=0.01875 m

=18.75 mm

in 2 significant figures, answer is 19 mm

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