Answer to Question #285946 in Classical Mechanics for Mukhtar

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 – (0.0625 cm/𝑠2)𝑡2. (a) Find the turtle’s initial velocity, initial position, and initial acceleration. (b) At what time t 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 10.00 cm from its starting point? What is the velocity at each of these times?

Answer 1

(a) We can find the turtle’s initial velocity, initial position, and initial acceleration from the kinematic equation:

$$x_f=x_i+v_it+\dfrac{1}{2}at^2,$$

here, $x_i$ is the initial position, $v_i$ is the initial velocity and $a$ is the initial acceleration.

Then, we get:

$$x_i=50\ cm,$$ $$v_i=2\ \dfrac{cm}{s},$$ $$\dfrac{1}{2}a=-0.0625\ \dfrac{cm}{s^2},$$ $$a=2\times(-0.0625\ \dfrac{cm}{s^2})=-0.125\ \dfrac{cm}{s^2}.$$

(b) We can find the time at which the velocity of the turtle zero from the kinematic equation:

$$v_f=v_i+at,$$ $$0=2\ \dfrac{cm}{s}-0.125\ \dfrac{cm}{s^2}t,$$ $$t=\dfrac{2\ \dfrac{cm}{s}}{0.125\ \dfrac{cm}{s^2}}=16\ s.$$

(c) When the turtle returns to its starting point $x_f=x_i.$ Therefore, we can rewrite our kinematic equation:

$$x_f-x_i=0=v_it+\dfrac{1}{2}at^2,$$ $$t(v_i+\dfrac{1}{2}at)=0.$$

The first root of this quadratic equation corresponds to the time when the turtle starts at starting point. The second one corresponds to the time when the turtle returns to the starting point:

$$t=-\dfrac{2v_i}{a}=-\dfrac{2\times2\ \dfrac{cm}{s}}{-0.125\ \dfrac{cm}{s^2}}=32\ s.$$

(d) Let's write the kinematic equation and find the times at which the turtle is at distance of 10.00 cm from its starting point:

$$60=50+2t-0.5\times0.125t^2,$$ $$0.0625t^2-2t+10=0.$$

This quadratic equation has two roots: $t_1=6.2\ s$, $t_2=25.8\ s$. So, we can accept both answers.

Also, we can write our kinematic equation as follows:

$$40=50+2t-0.5\times0.125t^2$$ $$0.0625t^2-2t-10=0.$$

This quadratic equation has two roots: $t_1=36.4\ s$, $t_2=-4.4\ s$. Since, the time can't be negative, we can accept only the first answer. Therefore, the times at which the turtle is at distance of 10.00 cm from its starting point are 6.2 s, 25.8 s and 36.4 s.

Finally, let's find the final velocity of turtle at each of these times:

$$v(t=6.2\ s)=2\ \dfrac{cm}{s}+(-0.125\ \dfrac{cm}{s^2})\times6.2\ s=1.22\ \dfrac{cm}{s},$$ $$v(t=25.8\ s)=2\ \dfrac{cm}{s}+(-0.125\ \dfrac{cm}{s^2})\times25.8\ s=-1.22\ \dfrac{cm}{s},$$ $$v(t=36.4\ s)=2\ \dfrac{cm}{s}+(-0.125\ \dfrac{cm}{s^2})\times36.4\ s=-2.55\ \dfrac{cm}{s}.$$