A stuntman drives a dirt bike on a curved track with a radius of 15.9 m. If he starts from rest and accelerates at 1.54 m/s2, at what time will the magnitude of the total acceleration of the bike be 5.29 m/s2? (Hint: the tangential acceleration will stay constant at 1.54 m/s2.)
Well, that stuntman on his dirt bike is really taking some ups and downs, huh? Let’s get cracking on this problem.
We know that the tangential acceleration will remain constant at 1.54 m/s². Now, we want to find the time when the magnitude of the total acceleration becomes 5.29 m/s².
Here's the deal: in circular motion, the total acceleration is the vector sum of the centripetal acceleration (directed towards the center of the circle) and the tangential acceleration.
Since we have the radius of the track (15.9 m), and the centripetal acceleration is given by the formula a_c = v^2 / r, and the tangential acceleration is constant at 1.54 m/s², we can solve this problem using some mathematical pizzazz.
We’ll start by finding the velocity v at the moment when the magnitude of the total acceleration is 5.29 m/s². The total acceleration squared is given by a_total^2 = v^2 / r^2 + a_t^2.
Plugging in the values, we have 5.29^2 = v^2 / (15.9)^2 + 1.54^2.
Squaring it out and rearranging gives us the equation: v^2 = (5.29^2 - 1.54^2) × (15.9)^2.
Now let's put on our math hats and calculate v by taking the square root of both sides of the equation.
Once we have v, we can use it to calculate the time it takes for the stuntman to reach that velocity using the formula v = u + at, where u is the initial velocity (0 m/s) and a is the tangential acceleration (1.54 m/s²).
So, setting v = u + at and solving for t, we can find the time.
I’m sorry if all this math makes your head spin faster than the stuntman on his bike, but it’s the way to calculate it. You got this!
To find the time at which the magnitude of the total acceleration of the dirt bike is 5.29 m/s^2, we need to consider the components of acceleration in the radial and tangential directions.
Given:
Radial acceleration (centripetal acceleration) = 5.29 m/s^2
Tangential acceleration = 1.54 m/s^2
Radius of the track = 15.9 m
The total acceleration can be calculated using the Pythagorean theorem:
Total acceleration = sqrt((Radial acceleration)^2 + (Tangential acceleration)^2)
Let's plug in the given values:
5.29 m/s^2 = sqrt((Radial acceleration)^2 + (1.54 m/s^2)^2)
To find the radial acceleration, we can use the formula:
Radial acceleration = (velocity)^2 / radius
Initially, the stuntman is at rest, so the initial velocity is 0 m/s.
5.29 m/s^2 = sqrt(((0 m/s)^2 / (15.9 m))^2 + (1.54 m/s^2)^2)
Simplifying further:
5.29 m/s^2 = sqrt((0)^2 + (1.54 m/s^2)^2)
Taking the square root on both sides:
5.29 m/s^2 = sqrt(0 + 2.3716 m^2/s^4)
Simplifying further:
5.29 m/s^2 = sqrt(2.3716 m^2/s^4)
Square both sides:
(5.29 m/s^2)^2 = (2.3716 m^2/s^4)
Simplifying further:
28.0241 m^2/s^4 = 2.3716 m^2/s^4
Now we can rearrange the equation for radial acceleration to solve for the velocity:
Radial acceleration = (velocity)^2 / radius
(28.0241 m^2/s^4) = (velocity)^2 / (15.9 m)
Cross-multiply:
(velocity)^2 = (28.0241 m^2/s^4) * (15.9 m)
Simplifying further:
(velocity)^2 = 446.04259 m^2/s^2
Taking the square root on both sides:
velocity = sqrt(446.04259 m^2/s^2)
velocity ≈ 21.12 m/s
Now we can use the formula for tangential acceleration to find the time:
Tangential acceleration = (final velocity - initial velocity) / time
1.54 m/s^2 = (21.12 m/s - 0 m/s) / time
Rearranging the equation:
time = (21.12 m/s) / 1.54 m/s^2
Simplifying:
time ≈ 13.71 seconds
Therefore, the magnitude of the total acceleration of the dirt bike will be 5.29 m/s^2 at approximately 13.71 seconds.
To find the time at which the magnitude of the total acceleration is 5.29 m/s², we need to consider the components of acceleration in this scenario.
Let's break down the acceleration into two components: tangential (at) and centripetal (ac). The tangential acceleration (at) is given as 1.54 m/s², and the centripetal acceleration (ac) is what we need to find.
The total acceleration (a) can be calculated using the following formula:
a = √(at² + ac²)
We know that at is 1.54 m/s² and a is 5.29 m/s². Plugging in these values, the equation becomes:
5.29 = √(1.54² + ac²)
Squaring both sides of the equation, we get:
5.29² = 1.54² + ac²
27.9841 = 2.3716 + ac²
Subtracting 2.3716 from both sides:
25.6125 = ac²
Now, let's solve for ac:
ac² = 25.6125
Taking the square root of both sides:
ac ≈ √(25.6125)
ac ≈ 5.0617 m/s²
So, the centripetal acceleration (ac) is approximately 5.0617 m/s².
Now that we know the centripetal acceleration, we can determine the time it takes for the magnitude of the total acceleration to reach 5.29 m/s².
Using the formula for centripetal acceleration:
ac = v² / r
where v is the velocity and r is the radius.
We can rearrange the equation to solve for v:
v = √(ac * r)
Plugging in the values:
v = √(5.0617 * 15.9)
v ≈ 8.523 m/s
Now, we can use the formula for acceleration:
a = (vf - vi) / t
where a is the acceleration, vf is the final velocity, vi is the initial velocity, and t is the time.
Since the stuntman starts from rest, the initial velocity (vi) is 0 m/s.
So, the equation becomes:
5.29 = (8.523 - 0) / t
Multiplying both sides of the equation by t:
5.29 * t = 8.523
Dividing both sides of the equation by 5.29:
t ≈ 1.611 seconds
Therefore, the magnitude of the total acceleration of the bike will be 5.29 m/s² at approximately 1.611 seconds.