changes in velocity

Question

At 10:15 am, Snail 1 crossed the 65cm mark with a velocity of 4cm/min. In a great burst of energy, snail 1 raced to the 90 cm mark in 3 minutes. His velocity as he crossed the 90 cm mark was 7 cm/min. what was the acceleration on snail 1?

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

Snail 3 trying to keep up with snail 2, managed to get to 80cm mark at 10:15am with a velocity of 6cm/min. Continuing his excellent effort, he crossed the finish line with a velocity of 7.2cm/min and gobbled the lettuce leaves at excatly 10:18am. What was the acceleration of the winning snail?

 

Please help.....

Answer 1

Acceleration is the change in velocity over the change in time = Δv/Δt.  To do these problems, you need to find out how much the speed changed and over what period of time it changed.

 

Snail 1 changes from 4 cm/min to 7 cm/min in 3 minutes.  Subtract the starting velocity (4 cm/min) from the ending velocity (7 cm/min) then divide by the time (3 min):

 

Snail 1 = (7 cm/min. - 4 cm/min)/(3 minutes) = ?    (remember to put down the units)

 

Snail 2 changed from 7 cm/min. down to 1 cm/min. in 3 minutes

 

Snail 2 = (1 cm/min. - 7 cm/min.)/(3 min.) = ?        (note that the acceleration is negative when you slow down)

 

Can you do snail 3?  Find the staring velocity, the ending velocity, and the time.