Answer:
a) stopping distance = (v^2 / 2*g*μ)
b) stopping distance = (v^2 / 2*g*μ)
Explanation:
We will use the constant acceleration formula
u^2 = v^2 + 2*a*s ------- (1)
- u is the final velocity (m/s)
- v is the initial velocity (m/s)
- a is the acceleration (m/s^2)
- s is the stopping distance (m/s)
Acceleration can be determined from the 2nd Law of motion
F = m*a ------- (2)
- F is the force to stop the car/truck (N)
- m is the mass of car (kg)
Coefficient of Friction is the ratio of applied force to the normal force, hence
μ = F/Fn ------- (3)
- Fn is the normal force (N)
Fn = m*g ------- (4)
- g is the gravitational acceleration (m/s^2)
Substituting equation (4) into equation (3), we get
F = m*g*μ ------- (5)
Substituting equation (5) in equation (2), we get
a = g*μ ------- (6)
Substituting equation (6) in equation (1), we get
u^2 = v^2 + 2*s*g*μ
Final velocity u is zero as the car is halting for part (a), hence
-v^2 = 2*s*g*μ
s = - ( v^2 / 2*g*μ) SI unit is meters, m
The negative sign can be ignored as the car is decelerating (negative acceleration), so
s = (v^2 / 2*g*μ)
For part (b), we will simply substitute 2*m instead of m in the equations and carry out the same procedure as part (a). The answer that we get is the same as part (a), that is,
s = (v^2 / 2*g*μ) SI unit is meters, m
Note: The reason we get the same answer for both the part is the fact that the stopping distance is independent of the mass. The force required to bring about truck is twice as much as that for the car but on the other hand the force generated due to friction in case of truck is also twice as much as that of the car. As all the other variables are similar, the stopping distance will be same.
