a) [tex]1.5\cdot 10^5 m/s^2[/tex]
b) 6.3 cm
c) 12.6 cm
Explanation:
a)
The acceleration of an object is the rate of change of its velocity; it is given by:
[tex]a=\frac{v-u}{t}[/tex]
where
u is the initial velocity
v is the final velocity
t is the time interval taken for the velocity to change from u to t
In this problem for the spore, we have:
u = 0 (the spore starts from rest)
v = 1.11 m/s (final velocity of the spore)
[tex]t=7.40\mu s = 7.40\cdot 10^{-6}s[/tex] (time interval in which the spore accelerates from zero to 1.11 m/s)
Substituting, we find the acceleration:
[tex]a=\frac{1.11-0}{7.40\cdot 10^{-6}}=1.5\cdot 10^5 m/s^2[/tex]
b)
Since the upward motion of the spore is a free fall motion (it is subjected to the force of gravity only), it is a uniformly accelerated motion (=constant acceleration, equal to the acceleration due to gravity: [tex]g=9.8 m/s^2[/tex]). Therefore, we can apply the following suvat equation:
[tex]v^2-u^2=2as[/tex]
where:
v = 0 is the final velocity of the spore (when it reaches the maximum height, its velocity is zero)
u = 1.11 m/s is the initial velocity (the velocity at which it is ejected)
[tex]a=-g=-9.8 m/s^2[/tex] is the acceleration (negative because it is downward)
s is the vertical displacement of the spore, which corresponds to the maximum height reached by the spore
Solving for s, we find:
[tex]s=\frac{v^2-u^2}{2a}=\frac{0^2-(1.11)^2}{2(-9.8)}=0.063 m = 6.3 cm[/tex]
c)
If the spore is ejected at a certain angle [tex]\theta[/tex] from the ground, then its motion is a projectile motion, which consists of two independent motions:
- A uniform horizontal motion, with constant horizontal velocity
- A uniformly accelerated motion along the vertical direction (free fall motion)
The horizontal range of a projectile, which can be derived from the equations of motion, is given by:
[tex]d=\frac{v^2 sin(2\theta)}{g}[/tex]
where
v is the initial velocity
[tex]\theta[/tex] is the angle or projection
g is the acceleration of gravity
From the equation, we observe that the maximum range is achevied when
[tex]\theta=45^{\circ}[/tex]
For this angle, the range is
[tex]d=\frac{v^2}{g}[/tex]
For the spore in this problem, the initial velocity is
v = 1.11 m/s
Therefore, the maximum range is
[tex]d=\frac{(1.11)^2}{9.8}=0.126 m = 12.6 cm[/tex]
