A particle moves according to a law of motion s = f(t), t ? 0, where t is measur

ID: 2829762 • Letter: A

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.

f(t) = te^(-t/4)

(a) Find the velocity at time t (in ft/s).

v(t) =

(b) What is the velocity after 3 s? (Round your answer to two decimal places.)

v(3) = ft/s

(d) When is the particle moving in the positive direction? (Enter your answer using interval notation.)

(e) Find the total distance traveled during the first 10 s. (Round your answer to two decimal places.)

(f) Find the acceleration at time t (in ft/s2).

a(t) =

Find the acceleration after 3 s. (Round your answer to three decimal places.)

a(3) = ft/s2

(h) When is the particle speeding up? (Enter your answer using interval notation.)

When is it slowing down? (Enter your answer using interval notation.)

given s=f(t) so we get v(t)=ds(t)/dt and a(t) = dv(t)/dt

so we get v(t) = e^(-t/4)- (t/4)e^(-t/4) = e^(-t/4)(1-t/4)

now a(t) = e^(-t/4)(1-t/4)(-1/4) + e^(-t/4)(-1/4) = -1/4[e^(-t/4)(1-t/4)]

a)solved above

b)t=3 so e^(-3/4)(1-3/4) = 0.118

d)when v(t)>0 so we get 1>t/4 and hence t<4 so t lies in the interval (-inf , 4)

e) s= te^(-t/4) t=10 so 10e^(-10/4) =

f)solved

-1/4[e^(-t/4)(1-t/4)] t=3 we get 0.0295

h) a>0 we get 1-t/4<0 => t>4 so (4,+inf) slowing in the the interval (-inf , 4)

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