The position in meters of a particle is given by f(t)= 14t-3t^2, where t is measured in seconds.
a) Evaluate f'(2) and interpret the results.
This is how I solved this. f'(t)=14-6t= 2(7-3t)
f'(2)=2(7-3(2))= 2 This means the velocity of the particle is 2 m/s when t=2.
b) On what intervals is the particle speeding up?
I don't know how to do this part.
Please help
f'=14-6t
correct on a)
b)
f'>0
14-6t>0
or t<14/6 so it is speeding up for all t less than 14/6
To determine the intervals when the particle is speeding up, you need to find the intervals where the velocity, given by f'(t), is increasing. Here's how you can solve it:
1. Find the derivative of the position function, f(t), to obtain the velocity function, f'(t):
f'(t) = 14 - 6t
2. Set this equation equal to zero and solve for t to find the critical points where the velocity is neither increasing nor decreasing:
14 - 6t = 0
6t = 14
t = 14/6
t = 7/3
3. Now, you want to determine the intervals where the velocity is increasing. To do this, you can pick a test point, let's say t = 0, and evaluate the sign of f'(t) for values less than and greater than the critical point (t = 7/3):
For t < 7/3: Choose t = 0:
f'(0) = 14 - 6(0) = 14
Since f'(0) is positive, the velocity is increasing for t values less than 7/3.
For t > 7/3: Choose t = 4:
f'(4) = 14 - 6(4) = -10
Since f'(4) is negative, the velocity is decreasing for t values greater than 7/3.
4. Therefore, the particle is speeding up on the interval (0, 7/3).