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

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.

Step-by-step 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 * 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=(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=(7^(2)- 2^(2))/(2(1))\\s=(45)/(2)\\ s=22.5m`

Therefore the distance of the snail is 22.5 m.