Consider the following where s is in feet and t is in seconds.

Question

s(t) = t³ + 4t² + 4t + 15

(a) Find v(t).

v(t)=?

(b) Find a(t).

a(t)=?

(c) Find v(3).

(d) Find a(3).

Answer 1

v(t) is the first derivative of s(t).

s(t) = t³ + 4t² + 4t + 15

v(t)= 3t^2+8t+4

v(3) = 3(3)^2+8(3)+4 =55

a(t) is the second derivative of s(t)

a(t)=6t+8

a(3)=6(3)+8=26

Answer 2

s(t) = t^3 + 4t^2 + 4t + 15 take the derivative to find v(t)

s = distance, v= velocity, a = acceleration

s'(t) = v(t) = 3t^2 +8t +4

take the derivative of v(t) to find a(t)

s"(t) = v'(t) = a(t) = 6t +8

v(3) = 3(3)^2 + 8(3) +4 = 27+24 + 4= 55 ft/sec

a(3) = 6(3) +8 = 26 ft/sec^2

(occasionally these problems also mention "jerk" which is the derivative of acceleration. In this example, jerk would be 6 ft/sec^3. If you're driving a car and increase your acceleration, you might feel a "jerk")