Q1. Please answer all the parts to this question A particle moves according to a law of motion s = f(t), t≥ 0, where t is measured in seconds and s in feet. (If an answer does not exist, enter DNE.)f(t) = t³ 9t² + 24t(a) Find the velocity (in ft/s) at time t.v(t) =(b) What is the velocity (in ft/s) after 1 second?v(1) =t =(c) When is the particle at rest? (Enter your answers as a comma-separated list.)(d) When is the particle moving in the positive direction? (Enter your answer using interval notation.)(e) Draw a diagram to illustrate the motion of the particle and use it to find the total distance (in ft) traveled during the first 6 seconds.fta(t) =(f) Find the acceleration (in ft/s2) at time t.ft/s²a(1) =ft/sFind the acceleration (in ft/s²) after 1 second.ft/s²ft/s(g) Graph the position, velocity, and acceleration functions for 0 ≤ t ≤ 6.3020a124510-10Sa6t30When is it slowing down? (Enter your answer using interval notation.)2010-10(h) When is the particle speeding up? (Enter your answer using interval notation.)Sa1456302010-10a12456A1245302010- 10a6

Note: Since you have posted a question with multiple subparts we will solve the first three subparts for you. To get the remaining subparts to be solved please post the complete question and specify the remaining subparts. A particle moves according to the law of a motion s = f(t), t≥0 where t is measured in seconds and s in feet. f(t)=t3-9t2+24t(a) Let f(t)=t3-9t2+24t To find the velocity at time t, differentiate f(t) with respect to t v(t)=ddtf(t)v(t)=ddtt3-9t2+24tv(t)=3t2-2(9t)+24v(t)=3t2-18t+24 Therefore, the velocity at time t is v(t)=3t2-18t+24 ft/s (b) The velocity after 1 second is v(1)=3(1)2-18(1)+24v(1)=3(1)-18+24v(1)=3-18+24v(1)=3+6v(1)=9 Therefore, v(1) = 9 ft/s (c) The particle is at rest when the velocity is zero. v(t)=03t2-18t+24=03t2-6t-12t+24=03tt-2-12(t-2)=03t-12t-2=0t-2=0 or 3t-12=0t=2 or 3t=12t=2 or t=123t=2 or t=4 Therefore, the particle is at rest at t=2, 4