A stuntman drives a dirt bike on a curved track with a radius of 17.0 m. If he starts from rest and accelerates at 1.74 m/s^2, at what time will the magnitude of the total acceleration of the bike be 7.04 m/s^2?
inward Ac = v^2/r
tangential a = dv/dt
7.04 =sqrt[a^2 + (v^2/r)^2 ]
49.56 = 3.03 + v^4/289
v = a t = 1.74 t
v^4 = 9.17 t^4
so
46.53 = 9.17 t^4/289
t^4 = 1466
t = 38.3 seconds
Check my arithmetic!!!
The WindSeeker ride at Canada's Wonderland lifts riders up to 300 feet into the air while spinning them around in a 12.2 m radius circle. The WindSeeker starts its circular motion from rest with a constant angular acceleration of 0.0174 rad/s2. It maintains that angular acceleration for 48.1 s, after which the angular speed remains constant for the rest of the ride - another 120. s. What is the top speed of the ride?
To find the time at which the magnitude of the total acceleration of the bike is 7.04 m/s^2, we can use the following steps:
1. Recall that the total acceleration of an object moving along a curved path can be calculated using the equation:
a_total = a_tangent + a_normal
Where:
- a_total is the total acceleration
- a_tangent is the component of acceleration tangent to the curved path
- a_normal is the component of acceleration normal (perpendicular) to the curved path
2. In this case, the given acceleration is 1.74 m/s^2. Since the acceleration is tangential to the curved path, we can write:
a_tangent = 1.74 m/s^2
3. The acceleration normal (a_normal) can be calculated using the centripetal acceleration formula:
a_normal = v^2 / r
Where:
- v is the velocity
- r is the radius of the curved path
4. In this question, the stuntman starts from rest, so the initial velocity (v) is 0 m/s. Therefore, the acceleration normal becomes:
a_normal = (0^2) / 17.0 m
Simplifying, we find that a_normal = 0 m/s^2.
5. Now we can use the equation for total acceleration to find the value of v when a_total = 7.04 m/s^2:
7.04 m/s^2 = 1.74 m/s^2 + 0 m/s^2
Therefore, v = 7.04 m/s^2.
6. Finally, we can use the equation for acceleration to find the time (t) taken to reach this velocity:
v = a * t
Rearranging, t = v / a.
Substituting the values, t = 7.04 m/s^2 / 1.74 m/s^2.
Calculating, we find that t ≈ 4.05 s.
Therefore, at approximately 4.05 seconds, the magnitude of the total acceleration of the bike will be 7.04 m/s^2.
To find the time at which the magnitude of the total acceleration of the bike is 7.04 m/s^2, we can use the equations of motion. Let's break down the problem step by step:
1. First, we need to determine the total acceleration of the bike at any given time. The total acceleration is the vector sum of the tangential acceleration (due to changing speed) and the centripetal acceleration (due to change in direction). Since the stuntman is driving on a curved track, the centripetal acceleration provides the curvature of the path.
The centripetal acceleration can be calculated using the formula:
Ac = v^2 / r
where Ac is the centripetal acceleration, v is the velocity, and r is the radius of the track.
2. Next, we need to calculate the initial velocity of the dirt bike at the starting point. The problem states that the stuntman starts from rest, so the initial velocity is zero (v0 = 0).
3. With the given acceleration and initial velocity, we can now find the time it takes for the magnitude of the total acceleration to reach 7.04 m/s^2. The magnitude of the total acceleration (A) can be calculated using the formula:
A = sqrt((at^2) + (Ac^2))
where at is the tangential acceleration.
By rearranging the equation to solve for time (t), we get:
t = sqrt((A^2 - Ac^2) / at^2)
Now, let's plug in the given values:
Radius of the track (r) = 17.0 m
Centripetal acceleration (Ac) = v^2 / r
Total acceleration (A) = 7.04 m/s^2
Tangential acceleration (at) = 1.74 m/s^2
Since the stuntman starts from rest, the initial velocity (v0) = 0.
Plug these values into the equation and solve for t:
Ac = v^2 / r
Ac = 0^2 / 17.0
Ac = 0 m/s^2
A = sqrt((7.04^2) + (0^2))
A = 7.04 m/s^2
Now we have the values to calculate the time:
t = sqrt((A^2 - Ac^2) / at^2)
t = sqrt((7.04^2 - 0^2) / (1.74^2))
t = sqrt(49.5616 / 3.0276)
t = sqrt(16.35)
So, the time it will take for the magnitude of the total acceleration to be 7.04 m/s^2 is approximately 4.04 seconds.