An 3000-kh car traveling at 70 m/s takes 8 m to stop under full braking. The same car under similar road conditions, traveling at 140 m/s, takes
so we are looking for the force of braking (ie, the max deacceleration allowed).
vf=vi+at
vf-vi=at
Force=mass*acceleration, so
acceleration=frictionforce/mass=mass*9.8N/kg*coefficentfriction/mass=9.8*coefficent
so if deacceleration is constant, regardless of speed.
and you Double (70 to 140) the change in speed, then time must be doubled (to 16seconds)
the braking force dissipates the kinetic energy of the vehicle
if the velocity of the vehicle is doubled , the KE is quadrupled ... KE = 1/2 m v^2
... so the brakes must dissipate four times the energy
if the braking force is constant , it takes four times as long to stop
To find the distance the car takes to stop under similar road conditions when traveling at 140 m/s, we can use the concept of proportionality.
We are given:
Initial speed of the car (u) = 70 m/s
Distance taken to stop at this speed (s1) = 8 m
We need to find:
Distance taken to stop when traveling at 140 m/s (s2)
According to the concept of proportionality, the stopping distance is directly proportional to the square of the initial speed. Therefore, we can set up the following equation:
s1/u1^2 = s2/u2^2
Substituting the given values:
8/(70)^2 = s2/(140)^2
Simplifying the equation:
8/4900 = s2/19600
Cross-multiplying:
8 * 19600 = 4900 * s2
Dividing both sides by 4900:
s2 = (8 * 19600) / 4900
Calculating the value of s2:
s2 = 32 m
Therefore, the same car traveling at 140 m/s takes 32 meters to stop under similar road conditions.
To answer this question, we can use the concept of deceleration and the equations of motion.
The deceleration (negative acceleration) of the car can be calculated using the formula:
deceleration = (final velocity - initial velocity) / time
In this case, we are given the initial velocity, final velocity, and time taken to stop.
Given data:
Initial velocity (u) = 140 m/s
Final velocity (v) = 0 m/s (since the car has stopped)
Time taken to stop (t) = ?
Using the formula:
deceleration = (v - u) / t
Substituting the known values:
deceleration = (0 - 140) / t
Now, let's solve for the deceleration.
Decrease = -140 / t (Since the final velocity is negative due to deceleration)
Next, we can use the equation of motion to find the distance traveled while decelerating:
distance = (initial velocity * time) + (1/2 * acceleration * time^2)
In this case, the initial velocity is 140 m/s, the time is unknown, and the distance traveled (d) is 8 m.
8 = (140 * t) + (1/2 * deceleration * t^2)
Since we have the value for deceleration from the previous equation, let's substitute it in:
8 = (140 * t) + (1/2 * (-140 / t) * t^2)
8 = 140t - 70t
Combine like terms:
8 = 70t
Divide both sides by 70:
t = 8 / 70
Now, calculate the value of t:
t ≈ 0.1143 seconds
Therefore, the same car under similar road conditions, traveling at 140 m/s, takes approximately 0.1143 seconds to stop.