Respuesta :
Answer:
The snail travel at the end of 5 s with a velocity of 12 m/s and the distance of the snail is 22.5 m.
Explanation:
Given that, the initial velocity of the snail is,
[tex]u=2m/s[/tex]
And the acceleration of the snail is,
[tex]a=1m/s^{2}[/tex]
And the time taken by the snail is,
[tex]t=5 sec[/tex]
Now according to first equation of motion,
[tex]v=u+at[/tex]
Here, u is the initial velocity, t is the time, v is the final velocity and a is the acceleration.
Now substitute all the variables
[tex]v=2m/s+ 1 \times 5 sec\\v=7m/s[/tex]
Therefore, the snail travel at the end of 5 s with a velocity of 7 m/s.
Now according to third equation of motion.
[tex]v^{2}- u^{2}=2as\\ s=\frac{v^{2}- u^{2}}{2a} \\[/tex]
Here, u is the initial velocity, a is the acceleration, s is the displacement, v is the final velocity.
Substitute all the variables in above equation.
[tex]s=\dfrac{7^{2}- 2^{2}}{2(1)}\\s=\dfrac{45}{2}\\ s=22.5m[/tex]
Therefore the distance of the snail is 22.5 m.
