Given s = 10 - 10t2 + 45t3, determine the velocity when t = 1 s.

velocity in m/s =

My Answer
S'= 0-20t + 135t^2
S'(1) = -20+135 = 115

velocity in m/s = 115

Can you please confirm this is correct.

Thanks!

S = 10 - 10t^2 + 45t^3.

S = 10 - 10*1 = 45*1 = 45 s.

NOTE: 1^2 = 1*1 = 1.
1^3 = 1*1*1 = 1.

OOPS! I didn't realize you were doing calculus.

Your calculus is correct.

To determine the velocity when t = 1 second, we need to find the derivative of the given function s with respect to t, denoted by s'(t).

The given equation is s = 10 - 10t^2 + 45t^3.

To find s'(t), we differentiate each term of the equation with respect to t:

The derivative of 10 with respect to t is 0 (since 10 is a constant).
The derivative of -10t^2 with respect to t is -20t.
The derivative of 45t^3 with respect to t is 135t^2.

Putting it all together, we have:

s'(t) = 0 - 20t + 135t^2.

Now, to find the velocity when t = 1 second, we substitute t = 1 into the derivative equation:

s'(1) = 0 - 20(1) + 135(1^2) = 0 - 20 + 135 = 115.

So, the velocity when t = 1 second is 115 m/s.

Therefore, your answer of 115 m/s is correct.