For a short period of time, the frictional driving force acting on the wheels of the 2.5-Mg van is N, where t is in seconds. If
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Answer:
top speed = 17.25
Total height = 281.19 m
Explanation:
given data
mass = 75 kg
thrust = 160 N
coefficient of kinetic friction = 0.1
solution
we get here frictional force acting that is
frictional force = .............1
frictional force = 0.1 × 75 × 9.8
frictional force = 73.5 N
and
Net force acting will be F = 160 - 73.5 N
F = 86.5 N
so
Acceleration in the First 15 second will be
F = ma .........2
86.5 = 75 × a
a = 1.15 m/s²
and
now After 15 second the velocity will be as
v = u + at ..........3
here u is 0
so v will be
V = 1.15 × 15
v = 17.25
and
now we get travels distance S in 15 s
s = u × t + 0.5 × a × t²
here u is 0
so distance s will be
s = 0.5 × a × t²
s = 0.5 × 1.15 × 15²
s = 129.37 m
and
now acceleration acting is
F =
m a =
a =
a = - 0.98
here it is negative it mean downward nature of acceleration
and
now we get distance s by this formula
V² - u² = 2 a s
here v velocity is 0 and
u initial velocity is 17.25 m/s
put here value
0 - 17.25² = 2 × (-0.98) × s
solve it we get
s = 151.82 m
so
Total height is
Total height = 129.37 m + 151.82 m
Total height = 281.19 m