Needs some help!!

**rugger15383** · January 27th 2010, 12:11 PM

Hi, could someone help me with this question?

A car reduces its velocity to standstill over a displacement of 500m with a retardation of 10 m/s². Calculate it's initial velocity

Im a bit stumped!

**ANDS!** · January 27th 2010, 05:35 PM

This should be in the calculus section.

Remember that the position function is defined as $s(t)=D$ , where D is distance. The derivative of the position funciton is velocity, $s'(t)=v(t)$ . And the second derivative of the position function, and the derivative of velocity is simply acceleration, $a(t)$ .

You are told that your vehicle is traveling at a constant deceleration of $-10\frac{m}{s^{2}}$ , which is the same as saying $a(t)=-10$ . Therefore, $v(t)=-10t+C$ , where C is our initial velocity. From the problem, we want to know what our initial velocity was, which means at some time "t", our velocity will be equal to zero (the car is at rest). In mathematical terms that means: $0=-10t+C \Rightarrow 10t=C$ .

So, we need to know what "t". From the equation for velocity, we see that $v(t)=-10t+C$ , and that the anti-derivative of this is $s(t)=-5t^{2}+Ct+D$ . But, we already know that C is 10t, so: $s(t)=-5t^{2}+10t^{2}+D$ . From the problem, we know that at some time "t", the exact same "t" that our final velocity is zero, that the car will have traveled 500 meters. Therefore: $500=-5t^{2}+10t^{2}+D$ . However, the cars initial position, relative to the problem, is 0, so we can say that D (which is initial position, in the same way that C is initial veloctiy) is equal to zero. Which leaves: $500=-5t^{2}+10t^{2} \Rightarrow 500=5t^{2}$ .

Can you take it from here? If you have any questions about the reasoning used feel free to ask.

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