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

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

Step-by-step 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+((-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(m)/(s) -2(m)/(s^(2) ) *(7.12s)=20.16(m)/(s)`
`V(driver)=34.4(m)/(s) -2(m)/(s^(2) ) *(21.63s)=-8.86(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