If an automobile of mass m rounds a curve, then its inward vector component of acceleration a N N is caused by the frictional force F of the road. Thus, it follows from the vector form of Newton’s second law [Equation (19)] that the frictional force and the normal component of acceleration are related by the equation F = m a N N . Thus, F = m k d s d t 2 Use this result to find the magnitude of the frictional force in newtons exerted by road on a 500 kg go-cart driven at a speed of 10km/h around a circular track or radius 15m. [ N o t e : 1 N = 1 kg . m / s 2 . ]
If an automobile of mass m rounds a curve, then its inward vector component of acceleration a N N is caused by the frictional force F of the road. Thus, it follows from the vector form of Newton’s second law [Equation (19)] that the frictional force and the normal component of acceleration are related by the equation F = m a N N . Thus, F = m k d s d t 2 Use this result to find the magnitude of the frictional force in newtons exerted by road on a 500 kg go-cart driven at a speed of 10km/h around a circular track or radius 15m. [ N o t e : 1 N = 1 kg . m / s 2 . ]
If an automobile of mass m rounds a curve, then its inward vector component of acceleration
a
N
N
is caused by the frictional force F of the road. Thus, it follows from the vector form of Newton’s second law [Equation (19)] that the frictional force and the normal component of acceleration are related by the equation
F
=
m
a
N
N
.
Thus,
F
=
m
k
d
s
d
t
2
Use this result to find the magnitude of the frictional force in newtons exerted by road on a 500 kg go-cart driven at a speed of 10km/h around a circular track or radius 15m.
[
N
o
t
e
:
1
N
=
1
kg
.
m
/
s
2
.
]
Quantities that have magnitude and direction but not position. Some examples of vectors are velocity, displacement, acceleration, and force. They are sometimes called Euclidean or spatial vectors.
A seasoned parachutist went for a skydiving trip where he performed freefall before deploying
the parachute. According to Newton's Second Law of Motion, there are two forcës acting on
the body of the parachutist, the forces of gravity (F,) and drag force due to air resistance (Fa)
as shown in Figure 1.
Fa = -cv
ITM EUTM FUTM
* UTM TM
Fg= -mg
x(t)
UTM UT
UTM /IM LTM
UTM UTM TUIM
UTM F UT
GROUND
Figure 1: Force acting on body of free-fall
where x(t) is the position of the parachutist from the ground at given time, t is the time of fall
calculated from the start of jump, m is the parachutist's mass, g is the gravitational acceleration,
v is the velocity of the fall and c is the drag coefficient. The equation for the velocity and the
position is given by the equations below:
EUTM PUT
v(t) =
mg
-et/m – 1)
(Eq. 1.1)
x(t) = x(0) –
Where x(0) = 3200 m, m = 79.8 kg, g = 9.81m/s² and c = 6.6 kg/s. It was established that the
critical position to deploy the parachutes is at 762 m from the ground…
Calc 3
I just wanted to make sure I am using the correct formula for the unit tangent vector. Does this appear to be correct to you guys? Thanks!
Precalculus: Mathematics for Calculus (Standalone Book)
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