Given:
u(initial velocity)=0
a=5.54m/s^2
v(final velocity)=2 m/s
v=u +at
Where v is the final velocity.
u is the initial velocity
a is the acceleration.
t is the time
2=0+5.54t
t=2/5.54
t=0.36 sec
Answer:less
Explanation:thats the answer
Answer:

Explanation:
mass of the bicycle + cyclist = 50 kg
constant speed = 6 km/h
a cyclist coasting down a 5.0° incline
the downward velocity is constant, so net acceleration must be zero
the air drag must be equal to gravitational force downward along the ramp
now for upward motion





Answer:
The hollow cylinder rolled up the inclined plane by 1.91 m
Explanation:
From the principle of conservation of mechanical energy, total kinetic energy = total potential energy

The total energy at the bottom of the inclined plane = total energy at the top of the inclined plane.

moment of inertia, I, of a hollow cylinder = ¹/₂mr²
substitute for I in the equation above;


given;
v₁ = 5.0 m/s
vf = 0
g = 9.8 m/s²

Therefore, the hollow cylinder rolled up the inclined plane by 1.91 m
Answer:
Explanation:
a )
While breaking initial velocity u = 62.5 mph
= 62.5 x 1760 x 3 / (60 x 60 ) ft /s
= 91.66 ft / s
distance trvelled s = 150 ft
v² = u² - 2as
0 = 91.66² - 2 a x 150
a = - 28 ft / s²
b ) While accelerating initial velocity u = 0
distance travelled s = .24 mi
time = 19.3 s
s = ut + 1/2 at²
s is distance travelled in time t with acceleration a ,
.24 = 0 + 1/2 a x 19.3²
a = .001288 mi/s²
= 2.06 m /s²
c )
If distance travelled s = .25 mi
final velocity v = ? a = .001288 mi / s²
v² = u² + 2as
= 0 + 2 x .001288 x .25
= .000644
v = .025 mi / s
= .0025 x 60 x 60 mi / h
= 91.35 mph .
d ) initial velocity u = 59 mph
= 86.53 ft / s
final velocity = 0
acceleration = - 28 ft /s²
v = u - at
0 = 86.53 - 28 t
t = 3 sec approx .