A particle moves along a straight line. Its position is given by y= squr of (t^2+7)?

Question

y is measured in feet and t is measured in seconds.

Find the acceleration of the particle when t=3

Please show all steps

y=square root of (t^2+7)

Answer 1

is that y=square of (t^2+7)
or square root of (t^2+7)?

y(t) is related to v(t) as
dy(t)/dt=v(t)
and
dv(t)/dt=a(t)

if y(t)=(t^2+7)^2
y(t)=t^4+14*t^2+49
v(t)=.25*t^3+7*t
a(t)=t^2/12+7
a(3)=9/12+7
a(3)=3/4+7=7.75

if it is square root of (t^2+7) just take the second derivative of
y(t)=(t^2+7)^(1/2)
then evaluate at t=3
using the chain rule:
dy(t)/dt=2*t/sqrt(t^2+7)
now use the quotient rule
f(x)=2*t
f'(x)=2
g(x)=sqrt(t^2+7)
g'(x)=2*t/sqrt(t^2+7)


the quotient rule is
D{f(x)/g(x)}=(g(x)*f'(x)-f(x)*g'(x))/
(g(x)^2)

so our second derivative is
[2*sqrt(t^2+7)-4*t^2/sqrt(t^2+7)]/
(t^2+7)

Evaluate at t=3
-0.625

j