2. A 5.2 x 10 kg car accelerates from rest under the

actions of two forces. One is a forward force of 3600

N provided by traction between the wheels and the

road. The other is 3350 N resistive force due to

various frictional forces. Determine how far the car

must travel for its speed to reach 7 m/s.

5.095m

Step-by-step explanation:

We first calculate the net force acting on the car

FN = Forward Force - Resistive Force

FN = 3 600 - 3350 = 250N

ACCORDING TO NEWTON'S SECOND LAW ACCELERATION IS GIVEN BY:

a = Fnet/m

=250N/5.2*10kg = 4.8077m/s^2

Since V = at

t = v/a 1

Equation of motion for the car;

S =ut + 1/2at^2

Where u = initial velocity for the car is zero

U=0

S=1/2at^2 2

Substitute to value of t in equation 1 into equation 2

S = 1/2 a(v/a)^2 = v^2/2a

= 7^2/2*4.8077=49/9.6154 = 5.095m

The car will travel for 5.095m

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step-by-step explanation:

point a is the center of the circle that passes through points e, f, and g and the center of the circle that passes through points x, y, and z.