To solve this problem, we need to use the physics concept of centripetal force, which is the force that keeps an object moving in a circular path. The formula for centripetal force is F = mv^2/r, where m is the mass of the object, v is the velocity (or speed), and r is the radius of the circular path.

In this case, the car must remain in contact with the speed bump, so the maximum force that can be exerted without the car losing contact is equal to the weight of the car, which is mass times gravity (F = mg).

Setting these two equations equal to each other gives us:

mv^2/r = mg

We can cancel out the mass (m) from both sides, and we're left with:

v^2/r = g

We're solving for v (the maximum speed), so we rearrange the equation to solve for v:

v = sqrt(gr)

Now we can plug in the given values. The radius (r) is 19.2 m, and the acceleration due to gravity (g) is approximately 9.8 m/s^2:

v = sqrt((9.8 m/s^2)(19.2 m)) = sqrt(188.16 m^2/s^2) = 13.71 m/s

So, the maximum speed that the car can have in order to remain in contact with the speed bump is approximately 13.71 m/s.

Question

To solve this problem, we need to use the physics concept of centripetal force, which is the force that keeps an object moving in a circular path. The formula for centripetal force is F = mv^2/r, where m is the mass of the object, v is the velocity (or speed), and r is the radius of the circular path.

In this case, the car must remain in contact with the speed bump, so the maximum force that can be exerted without the car losing contact is equal to the weight of the car, which is mass times gravity (F = mg).

Setting these two equations equal to each other gives us:

mv^2/r = mg

We can cancel out the mass (m) from both sides, and we're left with:

v^2/r = g

We're solving for v (the maximum speed), so we rearrange the equation to solve for v:

v = sqrt(gr)

Now we can plug in the given values. The radius (r) is 19.2 m, and the acceleration due to gravity (g) is approximately 9.8 m/s^2:

v = sqrt((9.8 m/s^2)(19.2 m)) = sqrt(188.16 m^2/s^2) = 13.71 m/s

So, the maximum speed that the car can have in order to remain in contact with the speed bump is approximately 13.71 m/s.

Knowee AI · Accepted Answer

To solve this problem, we need to use the physics concept of centripetal force, which is the force that keeps an object moving in a circular path. The formula for centripetal force is F = mv^2/r, where m is the mass of the object, v is the velocity (or speed), and r is the radius of the circular path.

In this case, the car must remain in contact with the speed bump, so the maximum force that can be exerted without the car losing contact is equal to the weight of the car, which is mass times gravity (F = mg).

Setting these two equations equal to each other gives us:

mv^2/r = mg

We can cancel out the mass (m) from both sides, and we're left with:

v^2/r = g

We're solving for v (the maximum speed), so we rearrange the equation to solve for v:

v = sqrt(gr)

Now we can plug in the given values. The radius (r) is 19.2 m, and the acceleration due to gravity (g) is approximately 9.8 m/s^2:

v = sqrt((9.8 m/s^2)(19.2 m)) = sqrt(188.16 m^2/s^2) = 13.71 m/s

So, the maximum speed that the car can have in order to remain in contact with the speed bump is approximately 13.71 m/s.

Questions 7 & 8 refer to the following in formation:A car of mass 760 kg travels along a flat road approaching a school crossing and hits a speed bump with a radius of 19.2 m.Question 7(3 marks)Calculate the maximum speed that the car can have in order to remain in contact with the speed hump.

Question

Solution

Similar Questions

Upgrade your grade with Knowee