I’m really just not sure how to type this into my calculator or even attempt to answer this question. 0113. (Calculator Active) For 0 ≤ t ≤ 2, a bug moves along a straight path with velocity v(t) = 2t cos(5 ln(t + 1)).-GRAPH(a) At what time(s) t does the bug turn around? Give a reason for your answer.*changes direction when V(+)=0O=2*cos(5)n(+1)(b) On what interval(s) does the bug have a negative acceleration?accel when concave downnegative2+ ((c) Find the velocity of the bug when a(t) = 3.Think(d) On what intervals is the bug speeding up? Give a reason for your answer.

Dear student, I will solve the first 3 subparts per the QnA guidelines. Please repost the remaining subpart separately. Given a bug that moves along a straight path with a velocity given as v(t)=2tcos(5 ln(t+1))...(i) for the given interval 0≤t≤2 Here the velocity is given as the function of time and has an exponential and a cosine term in it. For this, we need to answer the following.1) At what time t does the bug turn around? First, we plot the graph of our function v(t) a velocity-time graph for the interval 0≤t≤2. Now at the turning point, the velocity v should be zero. Hence in our v-t graph, the value of the v(t) coordinate should be zero and we see in our graph we have two points where the velocity is zero. Hence the value of t for which the velocity is zero is at t=0.37s and t=1.57s2) At what interval does the particle has a negative acceleration? Now for the interval where the particle has negative acceleration, the slope of the v-t graph will be negative. And, we can see from the graph the slope is negative from t=0s to t=0.98s0≤t≤0.98 We need to find the velocity of the bug when a(t)=3 We know the slop of the v-t graph denotes the acceleration so for a(t)=3the slope needs to be 3 i.e.dvdt=a(t)=3ddtcos(5 ln(t+1))Using the product rule we geta(t)≈2t·0.693cos(5log(t+1))-5sin(5log(t+1))t+1 Plotting the curve We can see a=3 for t=1.23s Hence we can get velocity by putting the value of t in eqn (i) v(t=1.23)=21.23cos(5 ln(1.23+1))v≈-1.52m/s

Answered: 3. (Calculator Active) For 0 ≤ t ≤ 2, a…