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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.

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  • 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}$$=t'(t)$ and $a=\large\frac{d^2s}{dt^2}=f''(t)$
  • When a particle starts from rest,velocity v and time t are 0. When a particle is thrown up, it reaches maximum height at which $v=0$ and then falls back to earth. When a moving particle comes rest, $v=0$
When the vehicle comes to a stop, velocity =0
Therefore $20-\large\frac{10}{3}$$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 $=60 \;m$


answered Jul 23, 2013 by meena.p

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