+322 The displacement (in meters) of a particle moving in a straight line is given by the equation of motion
\(s=\frac{4}{t^2}\), where t is measured in seconds. Find the velocity of the particle at t=a
+37146 The derivative of the displacement (or position) equation is the velocity function
(then the NEXT deriviative is the acceleration function......just an FYI)
s = 4/t^2 = 4 t^-2
s ' = v = -8 t^-3 sub in t = a
v at t= a = -8 a^-3 = -8/a^3
+129852 Write s as 4t^(-2)
The velocity is the derivative of the displacement function...so
s' (t) = -8t^(-3)
When t = a......the velocity is
s ' (a) = -8(a)^(-3)
+129852 OK....we have
[ f(t + h ) - f(t) ] / h =
[ 4 / (t + h)^2 - 4/t^2 ] / h =
[ 4t^2 - 4 ( t + h)^2 ] / [ h * t^2 * (t + h)^2 ] =
[ 4t^2 - 4t^2 - 8ht - 4h^2 ] / [ h * t^2 * ( t^2 + 2ht + h^2 ] =
[ h ( -8t - 4h) ] / [ h * ( t^4 + 2ht^3 + t^2h^2] .... [ divide out h ]
[ - 8t - 4h ] / [ (t^4 - 2ht^3 + t^2h^2 ] let h → 0 and we have that
[ -8t ] / [ t^4 ] =
-8 / t^3 = -8 t^(-3)
Don't worry....shortly.....you will learn to get to this result much faster !!!!
