Question

A driver of a car going 60 km/h suddenly sees the lights of a barrier 40 m ahead. It takes the driver 0.80 seconds before he applies the brakes, and the average acceleration during braking is -9.5 m/s. (A) Does the car hit the barrier? Explain. (B) What is the maximum speed at which the car could be moving and not hit the barrier 40 meters ahead?

Answer 1

a) The car doesn't hit the barrier because after he sees the lights of a barrier he only travels for 27.95m, enough to miss the barrier. b) The maximum speed at which the car could be moving and not hit the barrier 40 meters ahead is 21m/s or 75.6km/hr.

Step-by-step explanation:

a)

In order to solve this problem we must first do a drawing of the situation for us to visualize it better. (See picture attached).

As you may see on the drawing, the car still travels some distance when the driver notices the lights of the barrier. This distance is calculated for a constant velocity V:

x=Vt

the car has a velcity of 60km/h which is equivalent to:

`\frac {60km}{hr} * (1000m)/(1km)*(1hr)/(3600s)=16.667m/s`

so on the first 0.80s the car travels a distance of:

x=(16.667m/s)(0.8s)=13.333m

Next, the car breaks, so we can say it moves with a constant acceleration of -9.5m/s, so the distance is found by using the following formula:

`x=\frac {V_(f)^(2)-V_(0)^(2)}{2a}`

We know the final velocity will happen when the car stops, so the final velocity is zero, leaving us with:

`x=\frac {-V_(0)^(2)}{2a}`

so we can substitute the provided values to find the distance traveled by the car during this time.

`x=(-(16.667m/s)^(2))/(2(-9.5m/s^(2)))`

which yields:

x=14.62m

so in total the car traveled:

`x_(tot)=13.333m+14.62m = 27.95m`

Which is not enough for the car to hit the barrier.

b)

In order to solve this part of the problem, we must combine the two equations we got on the previous part to find a single equation that will represent the total displacement of the car:

`x=vt-(v^(2))/(2a)`

so we can now substitute the known values so we get:

`40=0.8v-(v^(2))/(2(-9.5))`

which simplifies to:

`40=0.8v+0.05263v^(2)`

this is a quadratic equation so it can be solved by using the quadratic formula, but first we must rewrite it in standard form, so we get:

`0.05263v^(2)+0.8v-40=0`

now we can use the quadratic formula:

`v=\frac{-b\pm\sqrt{b^(2)-4ac}}{2a}`

So we can substitute the given values:

`v=\frac{-0.8\pm\sqrt{(0.8)^(2)-4(0.05263)(-40)}}{2(0.05263)}`

which returns two answers:

v=21m/s and v=-36.20m/s

We take the positive answer since is the one that represents a moving towards the right side of the drawing.

So the maximum speed at which the car could be moving and not hit the barrier 40 meters ahead is 21m/s