Question

A super snail initially traveling at 2 m/s accelerates at 1 m/s^2 for 5 seconds. How fast will it be going at the end of the 5 seconds? How far did the snail travel?

Answer 1

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,

$u=2m/s$

And the acceleration of the snail is,

$a=1m/s^{2}$

And the time taken by the snail is,

$t=5 sec$

Now according to first equation of motion,

$v=u+at$

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

$v=2m/s+ 1 \times 5 sec\\v=7m/s$

Therefore, the snail travel at the end of 5 s with a velocity of 7 m/s.

Now according to third equation of motion.

$v^{2}- u^{2}=2as\\ s=\frac{v^{2}- u^{2}}{2a}$

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.

$s=\dfrac{7^{2}- 2^{2}}{2(1)}\\s=\dfrac{45}{2}\\ s=22.5m$

Therefore the distance of the snail is 22.5 m.