Questions: Question 13 The function f(t) gives the position of a body moving on a coordinate line, with s in meters and t in seconds. Time function: s = t^3 + 2t^2 - 24t, 0 ≤ t ≤ 2 Find the body's speed and acceleration at the end of the time interval. o 2 m/sec, -8 m/sec^2 o 6 m/sec, 0 m/sec^2 o 4 m/sec, -8 m/sec^2 o 2 m/sec, 8 m/sec^2

Solution Steps

The velocity \( v(t) \) is the first derivative of the position function \( s(t) \).

Given: \[ s(t) = t^3 + 2t^2 - 24t \]

Calculate the first derivative: \[ v(t) = \frac{d}{dt}(t^3 + 2t^2 - 24t) = 3t^2 + 4t - 24 \]

The acceleration \( a(t) \) is the first derivative of the velocity function \( v(t) \).

Calculate the first derivative of \( v(t) \): \[ a(t) = \frac{d}{dt}(3t^2 + 4t - 24) = 6t + 4 \]

Substitute \( t = 2 \) into the velocity and acceleration functions.

For velocity: \[ v(2) = 3(2)^2 + 4(2) - 24 = 3(4) + 8 - 24 = 12 + 8 - 24 = -4 \, \text{m/sec} \]

For acceleration: \[ a(2) = 6(2) + 4 = 12 + 4 = 16 \, \text{m/sec}^2 \]

Final Answer