 # The distince $x$ meters traveled by a vehical in time $t$ seconds after the brakes are applied is given by:$\;x=20 t-\large\frac{5}{3}$$t^{2}. Determine the distance the car travelled before it stops. Note: This is part 2nd of a 2 part question, split as 2 separate questions here. ## 1 Answer Comment A) Toolbox: • If s=f(t) is the distance function, representing the distance 's' travelled by a particle in time t, then the velocity and acceleration functions are v=\large\frac{ds}{dt}$$=t'(t)$ and $a=\large\frac{d^2s}{dt^2}=f''(t)$
• When a particle starts from rest,velocity v and time t are 0. When a particle is thrown up, it reaches maximum height at which $v=0$ and then falls back to earth. When a moving particle comes rest, $v=0$
$x=20t-\large\frac{5}{3}$$t^2 When the vehicle comes to a stop, velocity =0 Therefore 20-\large\frac{10}{3}$$t=0=>t=6 \;secs$
The distance travelled in $6 \;secs$ is obtained by substituting $t=6$ in (i)
$x=20 \times 6-\large\frac{5}{3}$$\times 36$
$\quad=120-60=60\; m$
The distance travelled before the vehicle comes to a half $=60 \;m$