A turtle crawls along a straight line, which we will call the x-axis with the po

Question

A turtle crawls along a straight line, which we will call the x-axis with the positive direction to the right. The equation for the turtle's position as a function of time is x(t) = 50.0 cm +(2.00 cm/s)t -(.0625 cm/s^2)t^2 At what time is the turtle third time a distance of 10.0 from its starting point?

Explanation / Answer

(a) x(t) = 50.0cm + (2.00 cm/s)t — (0.0625 cm/s2)t² x(0) = 50 cm v(t) = dx/dt = 2 cm/s - (2*0.0625 cm/s²)t = 2 cm/s - (0.125 cm/s²)t v(0) = 2 cm/s a(t) = dv/dt = -0.125 cm/s² = const a(0) = -0.125 cm/s² (b) v(t1) = 0 ==> 2 cm/s - (0.125 cm/s²)t1 = 0 ==> t1 = 2/0.125 = 16 s (c) x(t2) = 50 cm ==> 50.0cm + (2.00 cm/s)t2 — (0.0625 cm/s2)t2² = 50 (2.00 cm/s)t2 — (0.0625 cm/s2)t2² = 0 Divide the above equation by t2?0 ==> (2.00 cm/s) — (0.0625 cm/s2)t2 = 0 ==> t2 = 2/0.0625 = 32 s (d) x(t) = x(0) ± 10 cm x(t3) = x(0) + 10 = 50 + 10 = 60 ==> 50.0cm + (2.00 cm/s)t3 — (0.0625 cm/s2)t3² = 60 (0.0625 cm/s2)t3² - (2.00 cm/s)t3 - 10 = 0 Multiply the above equation by 16 ==> t3² - 32t3 - 160 = 0 Solve the quadratic to find t3 and t3'. Similarly for x(t4) = x(0) - 10 = 50 - 10 = 40 50.0cm + (2.00 cm/s)t4 — (0.0625 cm/s2)t4² = 40 I will continue, complete the answer and plot the graphs just in case you choose best answers to all of your correctly answered questions, you, ungrateful kid, who haven't chosen any best answer yet.

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