A 1500 kg car enters a section of curved road in the horizontal plane and slows down at a uniform rate from a speed of 100 km/h

Answer:

The question is incomplete; refer to the attachment for the diagram

Explanation:

Provided that,

The mass of the car

M = 1500kg

Entering the curve at Point A with a speed of

Va = 100km/hr = 100× 1000/3600

Va = 27.78m/s

The car slows down at a constant rate until it reaches point C at a speed of

Vc = 50km/hr = 50×1000/3600

Vc = 13.89m/s

Radius of curvature at point A

p = 400m

Radius of curvature at point B

p = 80m

The distance from point A to point B as indicated in the attachment is

S=200m

We need to determine the total horizontal forces at points A, B, and C applied by the road on the tire

The constant tangential acceleration can be computed using the equation of motion

Vc² = Va² + 2as

13.89² = 27.78² + 2 × a × 200

192.9 = 771.6 + 400a

400a = 192.9—771.6

400a = -578.7

a = -578.7 / 400

a = —1.45 m/s²

at = —1.45m/s²

The tangential acceleration is -1.45m/s² and it is negative due to the car decelerating

Since the vehicle is decreasing speed at a uniform rate, the tangential acceleration remains the same at every point

At point A

at = -1.45m/s²

At point B

at = -1.45m/s²

At point C

at = -1.45m/s²

Now,

We can compute the normal component of acceleration (centripetal acceleration) at each point since we know the radius of curvature

The centripetal acceleration can be calculated using

ac = v²/ p

At point A ( p = 400)

an = Va²/p = 27.78² / 400

an = 1.93 m/s²

At point B (p = ∞), since point B is the inflection point

Then,

an = Vb²/p =  Vb/∞ = 0

an = 0

At point C ( p = 80m)

an = Vc²/p = 13.89² / 80 = 2.41m/s²

an = 2.41 m/s²

Then,

The tangential force is

Ft = M•at

Ft = 1500 × 1.45

Ft = 2175 N.

Given that the tangential acceleration remains constant, this represents the tangential force at A, B, and C

Next, the normal force

At point A ( an = 1.93m/s²)

Fn = M•an

Fn = 1500 × 1.93

Fn = 2895 N

At point B (an=0)

Fn = M•an

Fn = 0 N

At point C (an= 2.41m/s²)

Fn = M•an

Fn = 1500 × 2.41

Fn = 3615 N.

Thus, the horizontal force acting at each point is

Using the right triangle vector method

F = √(Fn² + Ft²)

At point A

Fa = √(2895² + 2175²)

Fa = √13,111,650

Fa = 3621 N

At point B

Fb = √(0² + 2175²)

Fb = √2175²

Fb = 2175 N

At point C

Fc = √(3615² + 2175²)

Fc = √17,798,850

Fc = 4218.88 N

Fc ≈ 4219N