Answer:
68.46 m.
Explanation:
Given,
coefficient of friction,μ = 0.67
speed of truck, v = 30 m/s
distance travel by the truck to stop = ?
Now,
Calculation of acceleration
we know,
f = m a
and also
f = μ N = μ mg
Equating both the forces equation
m a = μ mg
a = μ g
a = 0.67 x 9.81
a = 6.57 m/s²
Now, using equation of kinematics
v² = u² + 2 a s
0 = 30² - 2 x 6.57 x s
s = 68.46 m
Hence, the minimum distance travel by the truck is equal to 68.46 m.
