Question

8. 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) = 5.0 cm + (4.00 cm/s) - (0.625 cm/s2)t2. (a) Find the turtle's initial velocity, initial position, and initial acceleration. (b) At what time is the velocity of the turtle zero? (c) How long after starting does it take the turtle to return to its starting point? (d) At what times t is the turtle a distance of 30.00 cm from its starting point? What is the velocity at each of these times?​

Answer 1

(a) The turtle's initial velocity is 4 m/s, initial position of the turtle is 5 cm, and initial acceleration is -1.25 m/s².

(b) The time when the velocity of the turtle is zero is 3.2 s.

(c) The time taken for the turtle to return to its starting point is 6.4 s.

(d) The time taken for the turtle to travel 30 cm is 0.08 s.

Initial velocity of the turtle

The initial velocity of the turtle is calculated as follows;

$v = \frac{dx}{dt} \\\\x = 5 + 4t -0.625t^2\\\\v(t) = 4 - 1.25t\\\\v(0) = 4-0\\\\v(0) = 4 \ m/s$

Initial acceleration of the turtle

The initial acceleration of the turtle is calculated as follows;

$a = \frac{dv}{dt} \\\\v(t) = 4 - 1.25t\\\\a = -1.25\ m/s^2$

Initial position of the turtle

x(t) = 5 + 4t - 0.625t²

x(0) = 5 cm

Time when the velocity becomes zero

v(t) = 4 - 1.25t

0 = 4 - 1.25t

1.25t = 4

t = 4/1.25

t = 3.2 s

Time taken to return to starting point

The total distance traveled is calculated as follows

v² = u² + 2ad

0 = (4)² + 2(-1.25)d

0 = 16 - 2.5d

2.5d = 16

d = 16/2.5

d = 6.4 m

Time to travel the given distance;

d = ut + ¹/₂at²

6.4 = (4)t + ¹/₂(-1.25)t²

6.4 = 4t - 0.625t²

0.625t² - 4t + 6.4 = 0

solve the quadratic equation using formula method;

t = 3.2 s

The time travel the distance two times, = 2 x 3.2 s = 6.4 s

Time taken for the turtle to travel 30 cm

d = ut + ¹/₂at²

0.3 = (4)t + ¹/₂(-1.25)t²

0.3 = 4t - 0.625t²

0.625t² - 4t + 0.3 = 0

solve the quadratic equation using formula method;

t = 0.08 s

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