Answer:
The question is incomplete, the complete question is "A car drives on a circular road of radius R. The distance driven by the car is given by d(t)= at^3+bt [where a and b are constants, and t in seconds will give d in meters]. In terms of a, b, and R, and when t = 3 seconds, find an expression for the magnitudes of (i) the tangential acceleration aTAN, and (ii) the radial acceleration aRAD3"
answers:
a.[tex]18a m/s^{2}[/tex]
b. [tex]a_{rad}=\frac{(27a +b)^{2}}{R}[/tex]
Explanation:
First let state the mathematical expression for the tangential acceleration and the radial acceleration.
a. tangential acceleration is express as
[tex]a_{tan}=\frac{d|v|}{dt} \\[/tex]
since the distance is expressed as
[tex]d=at^{3}+bt[/tex]
the derivative is the velocity, hence
[tex]V(t)=\frac{dd(t)}{dt}\\V(t)=3at^{2}+b\\[/tex]
hence when we take the drivative of the velocity we arrive at
[tex]a_{tan}=\frac{dv(t)}{dt}\\ a_{tan}=6at\\t=3 \\we have \\a_{tan}=18a m/s^2[/tex]
b. the expression for the radial acceleration is expressed as
[tex]a_{rad}=\frac{v^{2}}{r}\\a_{rad}=\frac{(3at^{2} +b)^{2}}{R}\\t=3\\a_{rad}=\frac{(27a +b)^{2}}{R}[/tex]
