v = v0 + a*t
a= (v-v0)/t
a=(7-4)/3
a=1 cm/min^2 SI units m/s
4 cm/min = 0,04/60=6,7 10^-4 m/s
7 cm/min = 0,001167
a=5*10^-4/180
a=2,76 *10^-6 m/s
a= (v-v0)/t
a=(7-4)/3
a=1 cm/min^2 SI units m/s
4 cm/min = 0,04/60=6,7 10^-4 m/s
7 cm/min = 0,001167
a=5*10^-4/180
a=2,76 *10^-6 m/s
Acceleration (a) = (Final velocity - Initial velocity) / Time
First, let's find the initial velocity of Snail 1 when it crossed the 65 cm mark. It is given that Snail 1 had a velocity of 4 cm/min at 10:15 am. Since we don't know when Snail 1 crossed the 65 cm mark, let's assume it took 't' minutes after 10:15 am to reach the 65 cm mark. Therefore, the initial velocity can be calculated as:
Initial velocity = 4 cm/min
Next, let's calculate the final velocity of Snail 1 when it crossed the 90 cm mark. It is given that Snail 1 took 3 minutes to cover the distance between 65 cm and 90 cm. Therefore, the final velocity can be calculated as:
Final velocity = Distance / Time = (90 cm - 65 cm) / 3 min = 25 cm / 3 min ≈ 8.33 cm/min
Now we have the initial velocity (4 cm/min), final velocity (8.33 cm/min), and the time (3 min). Plug these values into the acceleration formula:
Acceleration (a) = (Final velocity - Initial velocity) / Time
= (8.33 cm/min - 4 cm/min) / 3 min
= 4.33 cm/min / 3 min
≈ 1.44 cm/min²
Therefore, the acceleration of Snail 1 is approximately 1.44 cm/min².