What is the tangential acceleration (i.e., the acceleration parallel to the road) of car A?

SOMEONE ASKED 👇

What is the tangential acceleration (i.e., the acceleration parallel to the road) of car A?

HERE THE ANSWERS 👇

  • This problem has to do with the normal acceleration increasing/decreasing in the dips and bumps of the road. I would need the weight of the cars to get an exact answer so I will answer these in terms of the mass m.

    a) In this one we don’t need to worry about the normal acceleration since there is no curve.

    at = uk*Fn

    Fn = mg = 9.81m

    at = 1*9.81m = -9.81*m

    this is considered negative since the forward motion of the car is in the positive direction.

    b) Now we will find the normal acceleration caused by the curve.

    an = v^2/r = 27^2/200 = 0.135 m/s^2

    Fn = m*(an+ag) = (9.81+0.135)m = 9.945m

    at = uk*Fn = 1*9.945m = -9.945m

    c) Since the velocity and the radius are the same the normal acceleration is the same, however this time it is in the opposite direction of gravity.

    an = 0.135

    Fn = (9.81 – 0.135)m = 9.675m

    at = uk*Fn = -9.675m

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Your email address will not be published.

What is the tangential acceleration (i.e., the acceleration parallel to the road) of car A?

SOMEONE ASKED 👇

What is the tangential acceleration (i.e., the acceleration parallel to the road) of car A?

HERE THE ANSWERS 👇

  • This problem has to do with the normal acceleration increasing/decreasing in the dips and bumps of the road. I would need the weight of the cars to get an exact answer so I will answer these in terms of the mass m.

    a) In this one we don’t need to worry about the normal acceleration since there is no curve.

    at = uk*Fn

    Fn = mg = 9.81m

    at = 1*9.81m = -9.81*m

    this is considered negative since the forward motion of the car is in the positive direction.

    b) Now we will find the normal acceleration caused by the curve.

    an = v^2/r = 27^2/200 = 0.135 m/s^2

    Fn = m*(an+ag) = (9.81+0.135)m = 9.945m

    at = uk*Fn = 1*9.945m = -9.945m

    c) Since the velocity and the radius are the same the normal acceleration is the same, however this time it is in the opposite direction of gravity.

    an = 0.135

    Fn = (9.81 – 0.135)m = 9.675m

    at = uk*Fn = -9.675m

Similar Posts

Leave a Reply

Your email address will not be published.

What is the tangential acceleration (i.e., the acceleration parallel to the road) of car A?

SOMEONE ASKED 👇

What is the tangential acceleration (i.e., the acceleration parallel to the road) of car A?

HERE THE ANSWERS 👇

  • This problem has to do with the normal acceleration increasing/decreasing in the dips and bumps of the road. I would need the weight of the cars to get an exact answer so I will answer these in terms of the mass m.

    a) In this one we don’t need to worry about the normal acceleration since there is no curve.

    at = uk*Fn

    Fn = mg = 9.81m

    at = 1*9.81m = -9.81*m

    this is considered negative since the forward motion of the car is in the positive direction.

    b) Now we will find the normal acceleration caused by the curve.

    an = v^2/r = 27^2/200 = 0.135 m/s^2

    Fn = m*(an+ag) = (9.81+0.135)m = 9.945m

    at = uk*Fn = 1*9.945m = -9.945m

    c) Since the velocity and the radius are the same the normal acceleration is the same, however this time it is in the opposite direction of gravity.

    an = 0.135

    Fn = (9.81 – 0.135)m = 9.675m

    at = uk*Fn = -9.675m

Similar Posts

Leave a Reply

Your email address will not be published.

What is the tangential acceleration (i.e., the acceleration parallel to the road) of car A?

SOMEONE ASKED 👇

What is the tangential acceleration (i.e., the acceleration parallel to the road) of car A?

HERE THE ANSWERS 👇

  • This problem has to do with the normal acceleration increasing/decreasing in the dips and bumps of the road. I would need the weight of the cars to get an exact answer so I will answer these in terms of the mass m.

    a) In this one we don’t need to worry about the normal acceleration since there is no curve.

    at = uk*Fn

    Fn = mg = 9.81m

    at = 1*9.81m = -9.81*m

    this is considered negative since the forward motion of the car is in the positive direction.

    b) Now we will find the normal acceleration caused by the curve.

    an = v^2/r = 27^2/200 = 0.135 m/s^2

    Fn = m*(an+ag) = (9.81+0.135)m = 9.945m

    at = uk*Fn = 1*9.945m = -9.945m

    c) Since the velocity and the radius are the same the normal acceleration is the same, however this time it is in the opposite direction of gravity.

    an = 0.135

    Fn = (9.81 – 0.135)m = 9.675m

    at = uk*Fn = -9.675m

Similar Posts

Leave a Reply

Your email address will not be published.

What is the tangential acceleration (i.e., the acceleration parallel to the road) of car A?

SOMEONE ASKED 👇

What is the tangential acceleration (i.e., the acceleration parallel to the road) of car A?

HERE THE ANSWERS 👇

  • This problem has to do with the normal acceleration increasing/decreasing in the dips and bumps of the road. I would need the weight of the cars to get an exact answer so I will answer these in terms of the mass m.

    a) In this one we don’t need to worry about the normal acceleration since there is no curve.

    at = uk*Fn

    Fn = mg = 9.81m

    at = 1*9.81m = -9.81*m

    this is considered negative since the forward motion of the car is in the positive direction.

    b) Now we will find the normal acceleration caused by the curve.

    an = v^2/r = 27^2/200 = 0.135 m/s^2

    Fn = m*(an+ag) = (9.81+0.135)m = 9.945m

    at = uk*Fn = 1*9.945m = -9.945m

    c) Since the velocity and the radius are the same the normal acceleration is the same, however this time it is in the opposite direction of gravity.

    an = 0.135

    Fn = (9.81 – 0.135)m = 9.675m

    at = uk*Fn = -9.675m

Similar Posts

Leave a Reply

Your email address will not be published.

What is the tangential acceleration (i.e., the acceleration parallel to the road) of car A?

SOMEONE ASKED 👇

What is the tangential acceleration (i.e., the acceleration parallel to the road) of car A?

HERE THE ANSWERS 👇

  • This problem has to do with the normal acceleration increasing/decreasing in the dips and bumps of the road. I would need the weight of the cars to get an exact answer so I will answer these in terms of the mass m.

    a) In this one we don’t need to worry about the normal acceleration since there is no curve.

    at = uk*Fn

    Fn = mg = 9.81m

    at = 1*9.81m = -9.81*m

    this is considered negative since the forward motion of the car is in the positive direction.

    b) Now we will find the normal acceleration caused by the curve.

    an = v^2/r = 27^2/200 = 0.135 m/s^2

    Fn = m*(an+ag) = (9.81+0.135)m = 9.945m

    at = uk*Fn = 1*9.945m = -9.945m

    c) Since the velocity and the radius are the same the normal acceleration is the same, however this time it is in the opposite direction of gravity.

    an = 0.135

    Fn = (9.81 – 0.135)m = 9.675m

    at = uk*Fn = -9.675m

Similar Posts

Leave a Reply

Your email address will not be published.

What is the tangential acceleration (i.e., the acceleration parallel to the road) of car A?

SOMEONE ASKED 👇

What is the tangential acceleration (i.e., the acceleration parallel to the road) of car A?

HERE THE ANSWERS 👇

  • This problem has to do with the normal acceleration increasing/decreasing in the dips and bumps of the road. I would need the weight of the cars to get an exact answer so I will answer these in terms of the mass m.

    a) In this one we don’t need to worry about the normal acceleration since there is no curve.

    at = uk*Fn

    Fn = mg = 9.81m

    at = 1*9.81m = -9.81*m

    this is considered negative since the forward motion of the car is in the positive direction.

    b) Now we will find the normal acceleration caused by the curve.

    an = v^2/r = 27^2/200 = 0.135 m/s^2

    Fn = m*(an+ag) = (9.81+0.135)m = 9.945m

    at = uk*Fn = 1*9.945m = -9.945m

    c) Since the velocity and the radius are the same the normal acceleration is the same, however this time it is in the opposite direction of gravity.

    an = 0.135

    Fn = (9.81 – 0.135)m = 9.675m

    at = uk*Fn = -9.675m

Similar Posts

Leave a Reply

Your email address will not be published.

What is the tangential acceleration (i.e., the acceleration parallel to the road) of car A?

SOMEONE ASKED 👇

What is the tangential acceleration (i.e., the acceleration parallel to the road) of car A?

HERE THE ANSWERS 👇

  • This problem has to do with the normal acceleration increasing/decreasing in the dips and bumps of the road. I would need the weight of the cars to get an exact answer so I will answer these in terms of the mass m.

    a) In this one we don’t need to worry about the normal acceleration since there is no curve.

    at = uk*Fn

    Fn = mg = 9.81m

    at = 1*9.81m = -9.81*m

    this is considered negative since the forward motion of the car is in the positive direction.

    b) Now we will find the normal acceleration caused by the curve.

    an = v^2/r = 27^2/200 = 0.135 m/s^2

    Fn = m*(an+ag) = (9.81+0.135)m = 9.945m

    at = uk*Fn = 1*9.945m = -9.945m

    c) Since the velocity and the radius are the same the normal acceleration is the same, however this time it is in the opposite direction of gravity.

    an = 0.135

    Fn = (9.81 – 0.135)m = 9.675m

    at = uk*Fn = -9.675m

Leave a Reply

Your email address will not be published.

What is the tangential acceleration (i.e., the acceleration parallel to the road) of car A?

SOMEONE ASKED 👇

What is the tangential acceleration (i.e., the acceleration parallel to the road) of car A?

HERE THE ANSWERS 👇

  • This problem has to do with the normal acceleration increasing/decreasing in the dips and bumps of the road. I would need the weight of the cars to get an exact answer so I will answer these in terms of the mass m.

    a) In this one we don’t need to worry about the normal acceleration since there is no curve.

    at = uk*Fn

    Fn = mg = 9.81m

    at = 1*9.81m = -9.81*m

    this is considered negative since the forward motion of the car is in the positive direction.

    b) Now we will find the normal acceleration caused by the curve.

    an = v^2/r = 27^2/200 = 0.135 m/s^2

    Fn = m*(an+ag) = (9.81+0.135)m = 9.945m

    at = uk*Fn = 1*9.945m = -9.945m

    c) Since the velocity and the radius are the same the normal acceleration is the same, however this time it is in the opposite direction of gravity.

    an = 0.135

    Fn = (9.81 – 0.135)m = 9.675m

    at = uk*Fn = -9.675m

Leave a Reply

Your email address will not be published.