To anticipate a dip and a bump in the road the driver of a car applies her break to produce ga uniform deceleration . Her speed is 100km/hr at the bottom A of the dip and 50km/hr at the top C of the hump which is 120m along the road from A . If the passengers experiences a total acceleration of 3meter per second square at A . And if the radius of curvature at hump C is 150m . Calculate the radius of curvature of A . Calculate the acceleration at the point of inflexion B. find the acceleration at point C

Acceleration at c is 2.3 meter per seconds

To solve this problem, let's break it down step by step:

Step 1: Convert the speeds from km/hr to m/s.
To convert from km/hr to m/s, we divide the speed by 3.6 (since there are 3.6 seconds in an hour).
Speed at A: 100 km/hr * (1 hr/3600 s) * (1000 m/1 km) = 27.78 m/s
Speed at C: 50 km/hr * (1 hr/3600 s) * (1000 m/1 km) = 13.89 m/s

Step 2: Calculate the deceleration at point A.
Given that the passengers experience a total acceleration of 3 m/s^2 at point A, and the car is decelerating, we can write:
Acceleration at A = -deceleration at A = -3 m/s^2

Step 3: Calculate the radius of curvature at point C.
The radius of curvature, R, is related to the speed and acceleration through the formula:
R = (v^2) / a, where v is the speed and a is the acceleration.
Using the given values:
Radius of curvature at C = (13.89 m/s)^2 / 3 m/s^2 = 64.45 m

Step 4: Calculate the distance from A to B.
To find the distance from A to B, we need to know the position of the point of inflection, B. However, this information is not provided in the question. Without this information, we cannot find the distance from A to B or the acceleration at B.

Step 5: Calculate the acceleration at point C.
Given that the radius of curvature at C is 150 m, we can use the formula for acceleration in circular motion:
a = (v^2) / R, where v is the speed and R is the radius of curvature.
Acceleration at C = (13.89 m/s)^2 / 150 m = 1.28 m/s^2

Step 6: Calculate the radius of curvature at point A.
To calculate the radius of curvature at point A, we can use the formula for acceleration in circular motion:
a = (v^2) / R, where v is the speed and R is the radius of curvature.
Using the known acceleration at A (-3 m/s^2) and the speed at A (27.78 m/s):
-3 m/s^2 = (27.78 m/s)^2 / R_A
R_A = (27.78 m/s)^2 / (-3 m/s^2) = -256.63 m

Note: The negative sign indicates that the acceleration is in the opposite direction of the velocity.

To solve this problem, we can use the principles of circular motion and kinematics.

First, let's calculate the radius of curvature at point A.

1. Convert the speed at point A from km/hr to m/s:
Speed at A = 100 km/hr = (100 * 1000) / 3600 = 27.78 m/s

2. Use the formula for centripetal acceleration to calculate the total acceleration at point A:
Total acceleration at A = 3 m/s^2

The total acceleration at point A is the vector sum of the radial acceleration (ar) and the tangential acceleration (at):
Total acceleration at A = √(ar^2 + at^2)

Since there is no tangential acceleration at point A (as the car is moving in a straight line), the total acceleration is equal to the radial acceleration.

Therefore, ar = Total acceleration at A = 3 m/s^2

3. Use the formula for centripetal acceleration to calculate the radius of curvature at point A:
ar = (v^2) / r
r = (v^2) / ar
r = (27.78^2) / 3
r ≈ 248.33 meters

So, the radius of curvature at point A is approximately 248.33 meters.

Next, let's calculate the acceleration at the point of inflection, B.

The point of inflection represents the change from deceleration to acceleration, or vice versa. At this point, the acceleration is zero.

Now, given that point B is the point of inflection, we know that the acceleration at that point is zero.

Therefore, acceleration at B = 0 m/s^2

Finally, let's calculate the acceleration at point C.

The radius of curvature at point C is given to be 150 meters.

The total acceleration at C can be calculated using the same formula as before:
Total acceleration at C = √(ar^2 + at^2)

But since the car is at the top of the hump, the tangential acceleration is pointing opposite to the radial acceleration (which is directed towards the center of the curve).

So, the total acceleration at C = ar - at

Using the previous formula for centripetal acceleration:
ar = (v^2) / r

at = (v1^2 - v2^2) / 2s
where v1 is the initial speed at point C and v2 is the final speed at point C, and s is the distance between A and C.

Given:
v1 at C = 50 km/hr = (50 * 1000) / 3600 = 13.89 m/s
v at A = 27.78 m/s
s = 120 meters

ar = (27.78^2) / 150 ≈ 5.15 m/s^2

at = (13.89^2 - 27.78^2) / (2 * 120) ≈ -6.67 m/s^2

Total acceleration at C = ar - at = 5.15 - (-6.67) = 11.82 m/s^2

So, the acceleration at point C is approximately 11.82 m/s^2.