Problem
Particle A and Particle B begin moving away from a wall at the same time and are traveling in the same direction. The function $a(t)=7 t+14$ determines Particle A's distance from the wall (in $\mathrm{cm}$ ) given a number of $\mathrm{ms} t$ since the particle started moving. Similarly, $b(t)=6 t+10$ determines Particle B's distance from the wall (in $\mathrm{cm}$ ) given a number of ms $t$ since the particle started moving. a. Evaluate $a(24)$. \[ a(24)= \] Preview b. Evaluate $b(24)$. \[ b(24)= \] Preview c. Which particle was closer to the wall $24 \mathrm{~ms}$ after it started moving? Select an answer $\mathrm{V}$
Solution
Step 1 ：Substitute \(t=24\) into the function \(a(t)=7t+14\) to find the distance of Particle A from the wall after 24 ms.
Step 2 ：\[a(24)=7(24)+14=182\]
Step 3 ：Substitute \(t=24\) into the function \(b(t)=6t+10\) to find the distance of Particle B from the wall after 24 ms.
Step 4 ：\[b(24)=6(24)+10=154\]
Step 5 ：Compare the distances of Particle A and Particle B from the wall after 24 ms.
Step 6 ：Particle A is \(\boxed{182}\) cm away from the wall after 24 ms.
Step 7 ：Particle B is \(\boxed{154}\) cm away from the wall after 24 ms.
Step 8 ：Therefore, Particle B was closer to the wall 24 ms after it started moving.
