Question

Although ungraceful on land, walruses are fine swimmers. They normally swim at 1.94 m/s, and for short periods of time are capable of reaching speeds of nearly 9.72 m/s. Suppose a walrus accelerates from 1.95 m/s to 9.45 m/s over a distance of 95 m. What would be the magnitude of the walrus’s uniform acceleration?

Answer 1

Answer:

The magnitude of the walrus's uniform acceleration is $0.45m/s^{2}$

Explanation:

First, we need to gather the data the problem provides us with. We know the Walrus will start with a velocity of 1.95 m/s. This will be our initial velocity.

$V_{0}=1.95m/s$

Next, we know that in the end, the walrus will have a velocity of 9.45 m/s, this will be our final velocity.

$V_{f}=9.45m/s$

and finally, we know the walrus will travel 95m before he reaches the final velocity. This will be our displacement.

x= 95m

So we need to look for a formula we can use that contains: initial velocity, final velocity and displacement. Such a formula looks like this:

$a=\frac{V_{f} ^{2}-V_{0} ^{2}}{2x}$

so now we can plug the given data into the formula so we get:

$a=\frac{(9.45m/s) ^{2}-(1.95m/s)^{2}}{2(95m)}$

$a=\frac{89.3025m^{2}/s^{2} -3.8025m^{2}/s^{2}}{190m}$

$a=\frac{85.5m^{2}/s^{2}}{190m}$

which yields:

a=$0.45m/s^{2}$

So the magnitude of the walrus's uniform acceleration is $0.45m/s^{2}$