Snail 2, the speedy snail, flew past the 90-cm mark at 10:15 AM with a velocity of 7
cm/min. Unfortunately, Snail 2 stubbed his foot on the meterstick and could only limn
across the 93-cm mark 3 minutes later with a velocity of 1 cm/min. Calculate the
acceleration of Snail 2.

Respuesta :

The acceleration of the snail is [tex]-2 cm/min^2[/tex], or [tex]-5.6\cdot 10^{-6} m/s^2[/tex]

Explanation:

The acceleration of an object is given by the equation:

[tex]a=\frac{v-u}{\Delta t}[/tex]

where

v is the final velocity

u is the initial velocity

[tex]\Delta t[/tex] is the time taken for the velocity to change  from u to v

For the snail in this problem, we have:

u = 7 cm/min (initial velocity at the 90 cm mark)

v = 1 cm/min (final velocity at the 93 cm mark)

[tex]\Delta t = 3 min[/tex] (time interval between the snail being at the 90 cm mark and at the 93 cm mark)

Substituting into the equation, we find the acceleration of the snail:

[tex]a=\frac{1-7}{3}=-2 cm/min^2[/tex]

And converting into SI units,

[tex]a= -2 \frac{cm}{min^2} \cdot \frac{0.01 m/cm}{(60 sec/min)^2}=-5.6\cdot 10^{-6} m/s^2[/tex]

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