Question

A driver enters a one-lane tunnel at 34.4 m/s. The driver then observes a slow-moving van 154 m ahead travelling (in the same direction as the car) at a constant 5.65 m/s. The driver applies the brakes (ignore reaction time)of the car but can only accelerate at -2.00 m/s2 because the road is wet. How fast are you moving when you hit the rear of the van?
(A) 16.3 m/s
(B) 22 m/sec
(С) 4 m/sec
(D) 0 m/s

Answer 1

Answer:

Ans. B) 22 m/s (the closest to what I have which was 20.16 m/s)

Explanation:

Hi, well, first, we have to find the equations for both, the driver and the van. The first one is moving with constant acceleration (a=-2m/s^2) and the van has no acceletation. Let´s write down both formulas so we can solve this problem.

$X(van)=5.65t+154$

$X(driver)=34.4t+\frac{(-2)t^{2} }{2}$

or by rearanging the drivers equation.

$X(driver)=34.4t+t^{2}$

Now that we have this, let´s equal both equations so we can tell the moment in which both cars crashed.

$X(van)=X(driver)$

$5.65t+154=34.4t-t^{2}$

$0=t^{2} -(34.4-5.65)t+154$$0=t^{2} -28.75t+154$

To solve this equation we use the following formulas

$t=\frac{-b +\sqrt{b^{2}-4ac } }{2a}$

$t=\frac{-b +\sqrt{b^{2}-4ac } }{2a}$

Where a=1; b=-28.75; c=154

So we get:

$t=\frac{28.75 +\sqrt{(-28.75)^{2}-4(1)(154) } }{2(1)}=21.63s$$t=\frac{28.75 -\sqrt{(-28.75)^{2}-4(1)(154) } }{2(1)}=7.12s$

At this point, both answers could seem possible, but let´s find the speed of the driver and see if one of them seems ilogic.

$V(driver)=V_{0} +at$}

$V(driver)=34.4\frac{m}{s} -2\frac{m}{s^{2} } *(7.12s)=20.16\frac{m}{s}$$V(driver)=34.4\frac{m}{s} -2\frac{m}{s^{2} } *(21.63s)=-8.86\frac{m}{s}$

This means that 21.63s will outcome into a negative speed, for that reason we will not use the value of 21.63s, we use 7.12s and if so, the speed of the driver when he/she hits the van is 20.16m/s, which is closer to answer  A).

Best of luck