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 March 8th, 2009, 05:23 PM #1 Newbie   Joined: Jan 2009 Posts: 4 Thanks: 0 car problem 2) A car is traveling at 100 km/hr when the brakes are applied. If the brakes can give the car a constant negative acceleration of 8 m/s2, a) how long will it take the car to come to a stop, and b) how far will the car travel before stopping?
 March 8th, 2009, 06:40 PM #2 Newbie   Joined: Mar 2009 Posts: 3 Thanks: 0 Re: car problem No attempt at the problem...? This involves no calculus... this is just basic algebra and physics? Start by converting units... 100km/hr * 1000m/1km * 1hr / 60min * 1min/60sec = 27.77m/s Constant acceleration at 8m/s^2 It's easier to start with part b, in my opinion... Part B: v_F^2 = v_I^2 - 2 * a * d 0 = (27.77m/s)^2 - 2 * (-8m/s^2) * d (16m/s^2)d = (27.77m/s)^2 d = (27.77m/s)^2/(16m/s^2) = 48.22530864m Part A: d = -1/2*a*t^2 (48.22530864m) * 2 = -a * t^2 sqrt [(48.22530864m * 2)] / -a = t t = sqrt[(48.22530864m * 2)]/ -(-8m/s) = 3.472222222s
 March 11th, 2009, 06:12 PM #3 Member   Joined: Mar 2009 From: San Bernardino, California Posts: 50 Thanks: 0 Re: car problem Yes this problem can be done via Algebra but it can also be done using Calculus. First you are given that the deceleration due to brake application is $8 m/s^2$. If we define the function $a(t)$ as acceleration with respect to time we then have: $a(t)=-8$ Note however that the velocity of the car and the deceleration of the car use different units so first you must convert: $(100 km/h)(1000m/km)(1h/60min)(1min/60s)=(100*1000)/60^2)m/s=\frac{250}{9} m/s$ Continuing, by integrating the acceleration function we can get the function of the car's velocity, $v(t)$: $v(t)=\int{a(t)}dt=\int{-8}dt=-8t+C$ The initial velocity is when no time is passed, so $t=0$ and the initial velocity $v(0)=\frac{250}{9}$ is given and can be used to solve for $C$: $v(0)=\frac{250}{9}=-8(0)+C$ $C=\frac{250}{9}$ Therefore the function of velocity is then $v(t)=-8t+\frac{250}{9}$ From here the velocity must be zero for the car to be stopped so:  $t=\frac{125}{36}s\approx 3.47s$ A)The car has stopped after applying the brakes for 3.47 seconds. Similarly the rest of the problem follows that $r(t)=\int{v(t)}dt=\int{-8t+\frac{250}{9}}dt=-4t^2+\frac{250}{9}t+C$ In this case when no time has passed $t=0$ and $r(t)=0$ because all of these functions are relative to the initial conditions and the car's initial position and the car has not traveled any distance from its initial position, 0: $r(0)=0=-4(0)^2+\frac{250}{9}(0)+C$ $C=0$ $r(t)=-4t^2+\frac{250}{9}t$ Then at $t=\frac{125}{36}$ we know that the car has stopped and is its final position so: $r(\frac{125}{36})=-4(\frac{125}{36})^2+\frac{250}{9}(\frac{125}{36})= \frac{15625}{324}m\approx 48.23m$ B) The car will 48.23 meters before it comes to rest. Note that the answers hold true to the same answers derived by Tyler using algebraic based physics formulas.

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