A car is traveling at a speed of 15 m/s, slams on the brakes, locking the wheel of the car. The coefficient of kinetic friction between the tires and the road is 0.64. What is the speed of the vehicle after 2 seconds have passed?
Coefficient of kinetic friction =F/R=ma/mg
Which is a/g since the mass is equal
0.64=a/10 a=6.4m/s
NB. g=10m/s
Recall v=u+at
V=15+6.4×2
V-15=12.8
V=12.8+15
V=27.8m/s
To determine the speed of the vehicle after 2 seconds, we need to consider the deceleration caused by the kinetic friction between the tires and the road.
The formula to calculate the deceleration caused by friction is:
a = μ * g
where:
a is the acceleration (in this case, deceleration),
μ is the coefficient of kinetic friction, and
g is the acceleration due to gravity (which is approximately 9.8 m/s^2).
Given that the coefficient of kinetic friction (μ) is 0.64, we can substitute the values into the formula:
a = 0.64 * 9.8
a ≈ 6.272 m/s^2
Now, to find the final speed of the vehicle after 2 seconds, we can use the kinematic equation:
v = u + a * t
where:
v is the final velocity,
u is the initial velocity,
a is the acceleration (deceleration in this case), and
t is the time.
The initial velocity (u) is 15 m/s, the acceleration (a) is -6.272 m/s^2 (since it is deceleration), and the time (t) is 2 seconds. We can plug these values into the equation:
v = 15 + (-6.272) * 2
v = 15 - 12.544
v ≈ 2.456 m/s
Therefore, the speed of the vehicle after 2 seconds have passed is approximately 2.456 m/s.