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sweet [91]
3 years ago
7

A Thomson's gazelle can reach a speed of 13 m/s in 3.0 s. A lion can reach a speed of 9.5 m/s in 1.0 s. A trout can reach a spee

d of 2.8 m/s in 0.12 s. Which equation from the formula sheet should you use to determine which animal has the largest acceleration?
Physics
1 answer:
Bogdan [553]3 years ago
3 0

Answer:

The acceleration of Trout is largest.

Explanation:

Speed of Thomson's gazelle, v = 13 m/s in 3 s. Let us assumed that initial speed is 0. Acceleration is given by :

a_T=\dfrac{v-u}{t}\\\\a_T=\dfrac{v}{t}\\\\a_T=\dfrac{13\ m/s}{3\ s}\\\\a_T=4.34\ m/s^2

Speed of lion is 9.5 m/s in 10 s. Let us assumed that initial speed is 0. Acceleration is given by :

a=\dfrac{v-u}{t}\\\\a_l=\dfrac{v}{t}\\\\a_l=\dfrac{9.5\ m/s}{1\ s}\\\\a_l=9.5\ m/s^2

A trout can reach a speed of 2.8 m/s in 0.12 s. Let us assumed that initial speed is 0. Acceleration is given by :

a=\dfrac{v-u}{t}\\\\a_t=\dfrac{v}{t}\\\\a_t=\dfrac{2.8\ m/s}{0.12\ s}\\\\a_t=23.34\ m/s^2

It is clear that the acceleration of Thomson's gazelle, lion and trout is 4.34\ m/s^2, 9.5\ m/s^2\ and\ 23.34\ m/s^2 respectively. So, the acceleration of Trout is largest.

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<h2>Answer: 277.777 m</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told that the rock was<u> projected upward from the surface</u>, we will only use the equations related to the Y axis.

In this sense, the movement equations in the Y axis are:

y-y_{o}=V_{o}.t+\frac{1}{2}g.t^{2}    (1)

V=V_{o}-g.t    (2)

Where:

y  is the rock's final position

y_{o}=0  is the rock's initial position

V_{o}=30\frac{m}{s} is the rock's initial velocity

V is the final velocity

t is the time the parabolic movement lasts

g=1.62\frac{m}{s^{2}}  is the acceleration due to gravity at the surface of the moon

As we know y_{o}=0 , equation (2) is rewritten as:

y=V_{o}.t+\frac{1}{2}g.t^{2}    (3)

On the other hand, the maximum height  is accomplished when V=0:

V=V_{o}-g.t=0    (4)

V_{o}-g.t=0    

V_{o}=g.t    (5)

Finding t:

t=\frac{V_{o}}{g}    (6)

Substituting (6) in (3):

y=V_{o}(\frac{V_{o}}{g})+\frac{1}{2}g(\frac{V_{o}}{g})^{2}    (7)

y_{max}=\frac{{V_{o}}^{2}}{2g}    (8)  Now we can calculate the maximum height of the rock

y_{max}=\frac{{(30m/s)}^{2}}{(2)(1.62m/s^{2})}   (9)

Finally:

y_{max}=277.777m  

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