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Leokris [45]
3 years ago
6

Urgente

Physics
1 answer:
Naddik [55]3 years ago
4 0

Answer:

1) Maintain a smooth and progressive turning of the steering wheel

2) Increase the radius of the bend by moving along the outer circumference of the bend

3) Reduce the turning speed

4) Turn along banked portion of the road

Explanation:

Inertia is the tendency of an object to oppose a change in its velocity

When cornering sharply, the tendency is for the car to continue in its straight forward path, rather than the corner into which the driver is steering the wheels to

The inertia experienced by the car is related to the centrifugal force, given by the following relation;

F = m × v²/r

Where;

m = The mass of the car

v = The linear velocity of the car in the forward direction

r = The radius of the curve

Therefore, in order to overcome inertia when cornering sharply, the driver must do the following;

1) Maintain a smooth and progressive turning of the steering wheel, so as to gain traction control and reduce sudden changes in direction which may make the vehicle unbalanced as the inertia may be larger than the traction of the wheel or result in a large turning moment that may tip the car over

2) Increase the radius of the bend by moving along the outer circumference of the bend

Therefore;

i) Move along the left part of the lane when turning right

ii) Move along the right part of the lane when turning left

Moving in the manner described above, reduces the inertia, experienced while turning and improves the zone of vision of the car turning into the bend

3) Reduce the speed

The centrifugal force, which is proportional to the inertia, is directly proportional to the square of the velocity. Therefore, reducing the speed of the car, reduces the centrifugal force, and therefore, the inertia force experienced

Urgent

What must a driver do to overcome inertia when cornering sharply?

Safe driving cornering sharply

4) Ensure to turn along banked portion of the road, with banking angle θ given as follows;

θ = tan⁻¹(v²/(r·g))

Where;

g = The acceleration due to gravity

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