Can this even be solved??? help....?

Question

The driver of a car going 100.0 km/h suddenly sees the lights of a barrier 43.0 m ahead. It takes the driver 0.75 s to apply the brakes, and the average acceleration during braking is -10.0 m/s2.

a) What is the maximum speed at which the car could be moving and not hit the barrier 43.0 m ahead? Assume that the acceleration rate doesn't change.

_____km/h

Answer 1

here we have to basically calculate the distances covered during two time periods:

1.during braking(also known as reaction time)

2.after braking and before stopping.

for case 1 we proceed as follows:

during braking,we know that the driver will take some time in applying the brakes.now it means that he will cover some distance during the period of application.to find this distance,we have to know that during application,the car's speed does not change .it is only after application that the car's speed changes.to find the distance covered during braking,use the formula:displacement=velocity *time.time=0.75s.velocity=100km/h=27.78m...

therefore s1=0.75*27.78m/s

this gives us s1=20.835m

Now we have to calculate the speed of the car after the brakes have been applied

This is the max initial speed that the car will proceed with after the brakes have been applied and before the car comes to a stop

This is also the max speed at which the car can proceed safely and not hit the wall

We know that acceleration=10m/s^2

now the remaining distance that the car has to cover after braking = total distance  - distance coverd - during braking= 43m-20.835m=22.165m

using these values in the formula final velocity ^2= (initial velocity^2)+(2*acceleration*displacement...

we have final velocity^2= (27.78m/s)^2+ (2*-10m/s^2* 22.165m)

therefore final velocity = sqrt(328.4284m^2/s^2)=18.1225m/s

=65.241km/hr

this is the max speed at which the car can travel safely without hitting the wall

Answer 2

Yes, it can be solved and here is the solution.

Your working formula is

Vf^2  - Vo^2 = 2as   --- Call this Equation 1

where

Vf = final velocity = 0 (when the car finally stops)

Vo = velocity immediately before the brakes were applied

a = acceleration = 10 m/sec^2

s = distance travelled

First step is to check whether, given the conditions of the problem, the car will hit the barrier.

During the driver's reaction time = 0.75 sec., the car has already travelled,

x = VoT = (100 * 1000/3600)(0.75)

x = 20.83 meters

So, the effective remaining distance between the car and the barrier when the brakes were applied = 43 - 20.83 = 22.17 meters

Going back to Equation 1, substituting appropriate values,

0 - (100 * 1000/3600)^2 = 2(-10)(s)

NOTE the negative sign attached to the acceleration. This implies that the car was slowing down when the brakes were applied.

Solving for "s",

s = -(100 * 1000/3600)^2/(-20)

s = -771.60/-20

s = 38.58

Since the car will stop in 38.58 meters after the brakes were applied, it will definitely hit the barried because it was only 22.17 meters away from the barrier when the brakes were applied.

For the car not to hit the barrier, its stopping distance must be less than or equal to 22.17 meters.

Again, using Equation 1,

0 - Vo^2 = (2)(-10)(22.17)

Vo^2 = 20*22.17 = 443.40

Vo = 21.06 meters/sec = 75.8 km/hr.

ANSWER: At a speed of 75.8 kph, the car will not hit the barrier given the conditions of the problem.

Answer 3

To answer your main question, yes, it can be solved.