Question

at t = 0 a particle starts at rest and moves along a line in such a way that at time t its acceleration is 24t² feet per second per second. through how many feet does the particle move during the first 2 seconds? a 32 b 48 c 64 d 96 e 192

Explanation:

Step1: Find the velocity function

The acceleration function is \(a(t) = 24t^{2}\). Since \(v(t)=\int a(t)dt\), and \(v(0) = 0\) (starts at rest).

$$v(t)=\int24t^{2}dt=24\times\frac{t^{3}}{3}+C = 8t^{3}+C$$

Substitute \(t = 0\), \(v(0)=0\), so \(C = 0\). Then \(v(t)=8t^{3}\).

Step2: Find the position function

Since \(s(t)=\int v(t)dt\) and \(s(0) = 0\) (initial position).

$$s(t)=\int8t^{3}dt=8\times\frac{t^{4}}{4}+D=2t^{4}+D$$

Substitute \(t = 0\), \(s(0) = 0\), so \(D = 0\). Then \(s(t)=2t^{4}\).

Step3: Calculate the displacement at \(t = 2\)

Substitute \(t = 2\) into \(s(t)\):

$$s(2)=2\times2^{4}=2\times16 = 32$$

Answer:

A. 32