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.