Question

1. The acceleration of a particle in rectilinear motion is defined by a = k square root of v , where a is in m/s^2 , v is is m/s and k is a constant. Given that at times t = 2 sec and t = 3 sec, the velocities are respectively 4 m/s and 9 m/s, and the displacement at t = 3 sec is 20 m. Determine the values of k and Vo or Vinitial. Write the equation of motion.

2. A car starts from rest at Point O. A car covers 100 m in 10 seconds (Point A to B), while accelerating uniformly at a rate of 1 m/s^2. Determine
a) Velocities of the car at Point A and Point B.
b) Distance traveled before coming to this point A assuming it started from rest
c) Its velocity after the next 10 seconds (Point C)

Answer 1

Answer on Question #72505-Physics-Other

1. The acceleration of a particle in rectilinear motion is defined by $a = k$ square root of $v$, where $a$ is in $m/s^2$, $v$ is $m/s$ and $k$ is a constant. Given that at times $t = 2$ sec and $t = 3$ sec, the velocities are respectively $4 \, \text{m/s}$ and $9 \, \text{m/s}$, and the displacement at $t = 3$ sec is $20 \, \text{m}$. Determine the values of $k$ and $V_o$ or Vinitial. Write the equation of motion.

Solution

$$a = \frac{dv}{dt} = k\sqrt{v}$$ $$\frac{dv}{\sqrt{v}} = kt$$ $$\frac{\sqrt{v}}{\frac{1}{2}} = kt + C$$ $$2(\sqrt{v} - \sqrt{v_0}) = kt$$ $$\sqrt{v} = \sqrt{v_0} + \frac{kt}{2}$$ $$v = \left(\sqrt{v_0} + \frac{kt}{2}\right)^2$$ $$4 = (\sqrt{v_0} + k)^2$$ $$9 = \left(\sqrt{v_0} + \frac{3k}{2}\right)^2$$ $$\sqrt{v_0} + k = 2$$ $$\sqrt{v_0} + \frac{3k}{2} = 3$$ $$\frac{k}{2} = 1$$ $$k = 2\frac{m^2}{\frac{3}{s^2}}.$$ $$\sqrt{v_0} + 2 = 2$$ $$v_0 = 0\frac{m}{s}.$$ $$v = \left(\frac{2t}{2}\right)^2 = t^2.$$ $$\frac{dx}{dt} = v = t^2$$ $$x = C + \frac{t^3}{3}$$ $$20 = C + \frac{3^3}{3} = C + 9$$ $$C = 11.$$

The equation of motion is

$$x = 11 + \frac{t^3}{3}$$

2. A car starts from rest at Point O. A car covers $100\,\mathrm{m}$ in 10 seconds (Point A to B), while accelerating uniformly at a rate of $1\,\mathrm{m/s^2}$. Determine

a) Velocities of the car at Point A and Point B.

b) Distance traveled before coming to this point A assuming it started from rest

c) Its velocity after the next 10 seconds (Point C)

**Solution**

a)

$$d = a \frac{t^2}{2}$$ $$d_2 - d_1 = \frac{a}{2} (t_2^2 - t_1^2) = \frac{a}{2} (t_2 - t_1) (t_2 + t_1)$$ $$100 = \frac{1}{2} (10) (t_2 + t_1)$$ $$(t_2 + t_1) = 20$$ $$t_2 - t_1 = 10 \rightarrow t_2 = t_1 + 10.$$ $$(t_1 + 10 + t_1) = 20$$ $$2t_1 = 10$$ $$t_1 = 5\,\mathrm{s}.$$ $$t_2 = 15\,\mathrm{s}.$$ $$v_a = 1(5) = 5\,\frac{\mathrm{m}}{\mathrm{s}}.$$ $$v_b = 1(15) = 15\,\frac{\mathrm{m}}{\mathrm{s}}.$$

b)

$$d _ {1} = \frac {1}{2} (5) ^ {2} = 12.5 \, \mathrm{m}.$$

c)

$$v _ {c} = 1 (15 + 10) = 25 \, \frac{\mathrm{m}}{\mathrm{s}}.$$

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