Answer: At time 18.33 seconds it will have moved 500 meters.
Explanation:
Since the acceleration of the car is a linear function of time it can be written as a function of time as
[tex]a(t)=5(1-\frac{t}{15})[/tex]
[tex]a=\frac{d^{2}x}{dt^{2}}\\\\\therefore \frac{d^{2}x}{dt^{2}}=5(1-\frac{t}{15})[/tex]
Integrating both sides we get
[tex]\int \frac{d^{2}x}{dt^{2}}dt=\int 5(1-\frac{t}{15})dt\\\\\frac{dx}{dt}=v=5t-\frac{5t^{2}}{30}+c[/tex]
Now since car starts from rest thus at time t = 0 ; v=0 thus c=0
again integrating with respect to time we get
[tex]\int \frac{dx}{dt}dt=\int (5t-\frac{5t^{2}}{30})dt\\\\x(t)=\frac{5t^{2}}{2}-\frac{5t^{3}}{90}+D[/tex]
Now let us assume that car starts from origin thus D=0
thus in the first 15 seconds it covers a distance of
[tex]x(15)=2.5\times 15^{2}-\farc{15^{3}}{18}=375m[/tex]
Thus the remaining 125 meters will be covered with a constant speed of
[tex]v(15)=5\times 15-\frac{15^{2}}{6}=37.5m/s[/tex]
in time equalling [tex]t_{2}=\frac{125}{37.5}=3.33seconds[/tex]
Thus the total time it requires equals 15+3.33 seconds
t=18.33 seconds
