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Two stunt drivers drive their cars along a straight horizontal road. The first car is travelling at $30$ m/s and is followed by the second car, $16$ m behind it, both cars are travelling with equal speeds. At an instant the driver of the first car applies the brakes decelerating at $3$ m/s/s. Two seconds later , the second car brakes and decelerates at $4$ m/s/s. The time it takes the cars to collide?. $$s1: 30t-1.5t^2+16$$ $$s2:30(t+2)-2(t+2)^2$$ $s1=s2 -b$ formula and $t=3.66$ rounded to 2 decimal places Just not sure of my answer can someone plz check my answer, Thank u.

Philip O'Neill
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  • You should show the steps you went through. It is easier to check work than reproduce it, and if you have made an error we can identify it. – Ross Millikan Feb 10 '15 at 22:43
  • You should not have $t+2$ in either of the places you have it. It says that whatever formula you plug it into has been acting for two seconds longer than the time elapsed since the first car started braking. – David K Feb 10 '15 at 22:51
  • so should it be 30(t+2)-2t^2 – Philip O'Neill Feb 10 '15 at 22:52
  • No. See what happens at $t=0$, for example: your formula says the second car is at $60$ while the first is at $16$--how did the second car end up ahead? – David K Feb 10 '15 at 22:54
  • Also look at what the second car is doing. For $2$ seconds it moves at constant speed. Then afterwards it moves at a decreasing speed (decelerating). You cannot express both kinds of motion in a single formula like $s=s_0 + vt + \frac12 at^2.$ You'll need one calculation to say what happens for the first two seconds, another one to say what happens afterward. – David K Feb 10 '15 at 22:59
  • Of course you can $s=s_0+vt+1/2a{ONE}(t-2)^2$, where $ONE(t)$ is 1 for $t>0$ and zero for $t<0$. – Alexander Vigodner Feb 11 '15 at 18:00

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Equation let $d_1, d_2$ be the distance the cars 1 and 2 pass before the collision. $d_1=30t-3t^2/2$, $d_2=30t-4(t-2)^2/2$ We know that $d2=d1+16$ So we get the following equation $$ 30t-4(t-2)^2/2=16+30t-3t^2/2 $$ So regrouping: $$ -2t^2-8+8t=16-3t^2/2 \to ~ t^2+48-16t=0 $$ The solution $t=4$ and $12$, but the feasible solution obviously is the first one. So $t=4s$ Remark : if u want to know the time since the second car starts the deacceleration then $t=t-2=2$

The simplest check. The fist car starts to slow down. For 2 sec it passes $60-3*2^2/2=54 m$ The second car passes $60m$.For the next 2 sec the first car passes $24*2-3 2^2/2=42m$. The second car passes $30*2-4*2^2/2=52m$. So the first car passes $96$ and the second car:$112=96+16$.

Alexander Vigodner
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