Question

Given p(t)=−0.077t³+2.11t²−7.1t+166, find the acceleration at

t=2.
a) 4.264m/s²
b) -4.264m/s²
c) 12.792m/s²
d)-12.792m/s²

Answer 1

To find the acceleration at t = 2, we differentiate the function p(t) twice. First, we find the derivative of p(t), which is v(t). Then, we find the derivative of v(t), which is a(t). Finally, we substitute t = 2 into a(t) to find the acceleration, which is approximately 3.296 m/s².

Step-by-step explanation:

To find the acceleration at t = 2, we need to differentiate the function p(t) twice. First, we find the derivative of p(t), which is v(t). Then, we find the derivative of v(t), which is a(t).

So, let's start by finding the derivative of p(t):

p'(t) = (-0.077t^3 + 2.11t^2 - 7.1t + 166)'

p'(t) = -0.231t^2 + 4.22t - 7.1

Next, let's find the derivative of v(t):

v'(t) = (-0.231t^2 + 4.22t - 7.1)'

v'(t) = -0.462t + 4.22

Finally, let's substitute t = 2 into a(t) to find the acceleration:

a(2) = -0.462(2) + 4.22

a(2) = -0.924 + 4.22

a(2) = 3.296

Therefore, the acceleration at t = 2 is approximately 3.296 m/s².