A test car starts out from rest on a horizontal circular track of 80m radius and increases it's speed at a uniform rate to reach 27.78 m/s in 10s. Find the magnitude of the total acceleration of the car 8s after the start.
Ok so I know that the total acceleration is the square root of the tangential acceleration squared plus the normal acceleration squared. What I'm not sure of is how to find these components at t = 8s, since the velocity isn't constant then. It says that it increases it's speed at a constant rate to reach the 27.78 m/s so does that mean that the acceleration is the same for any time during that 10s interval? Any help would be appreciated
The radial accleration is V^2/R, where V is the velocity at that time. At t=8 s, V would be 4/5 of 27.78 m/s.
The tangential acceleration is the tate at which the speed increases with time, which is 2.778 m/s^2
thank you so much! this really helped me see where i went wrong
To find the magnitude of the total acceleration of the car 8 seconds after the start, we can break it down into tangential and normal components.
We know that the car starts from rest and increases its speed at a uniform rate to reach 27.78 m/s in 10 seconds. This means that the acceleration is constant during this interval.
To find the acceleration at t = 8s, we can use the equation for uniform acceleration, which states that:
v = u + at
Where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Given that the initial velocity (u) is 0 m/s (since the car starts from rest), the final velocity (v) is 27.78 m/s, and the time (t) is 10s, we can rearrange the equation to solve for acceleration (a):
a = (v - u) / t
a = (27.78 - 0) / 10
a = 2.78 m/s²
So, we have found that the acceleration during the 10s interval is 2.78 m/s².
Now, we need to find the velocity at t = 8s. We can use the equation:
v = u + at
Given that u is 0 m/s, a is 2.78 m/s², and t is 8s, we can calculate v:
v = 0 + (2.78 * 8)
v = 22.24 m/s
Now, we have the tangential component of velocity at t = 8s, which is 22.24 m/s.
To find the normal acceleration (which points towards the center of the circular track), we can use the equation:
a_normal = v² / r
Where v is the tangential velocity and r is the radius of the circular track.
Given that v is 22.24 m/s and r is 80m, we can calculate a_normal:
a_normal = (22.24)² / 80
a_normal = 6.17 m/s²
Now, we have the normal acceleration at t = 8s, which is 6.17 m/s².
Finally, we can calculate the magnitude of the total acceleration using the equation you mentioned:
a_total = sqrt(a_tangential² + a_normal²)
a_total = sqrt(2.78² + 6.17²)
a_total = sqrt(7.7284 + 38.0889)
a_total = sqrt(45.8173)
a_total = 6.77 m/s²
Therefore, the magnitude of the total acceleration of the car 8 seconds after the start is 6.77 m/s².
To find the magnitude of the total acceleration of the car 8 seconds after the start, we need to find both the tangential acceleration and the normal acceleration at that time.
First, let's find the tangential acceleration. Tangential acceleration represents the rate at which the speed (or magnitude of velocity) of the car is changing.
Given that the car starts from rest and reaches a speed of 27.78 m/s in 10 seconds, we can use the formula for constant acceleration:
v = u + at,
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
At t = 10s, v = 27.78 m/s and u = 0 m/s:
27.78 = 0 + a * 10,
a = 27.78 / 10,
a = 2.78 m/s^2.
Now we can find the tangential acceleration at t = 8s using the same formula:
v = u + at,
where v = 27.78 m/s, u = 0 m/s, a = 2.78 m/s^2, and t = 8s:
27.78 = 0 + 2.78 * 8,
27.78 = 22.24,
a = (27.78 - 0) / 8,
a = 3.47 m/s^2.
Next, let's find the normal acceleration. Normal acceleration represents the rate at which the direction of velocity is changing, which is always directed toward the center of the circular path.
Since the car is moving on a circular track, the normal acceleration can be calculated using the centripetal acceleration formula:
a_n = v^2 / r,
where a_n is the normal acceleration, v is the magnitude of velocity, and r is the radius of the circular track.
At t = 8s, v = 27.78 m/s and r = 80m:
a_n = (27.78)^2 / 80,
a_n = 9.61 m/s^2.
Finally, to find the magnitude of the total acceleration, we can use the Pythagorean theorem:
a_total = sqrt(a_t^2 + a_n^2),
where a_t is the tangential acceleration and a_n is the normal acceleration.
At t = 8s, a_t = 3.47 m/s^2 and a_n = 9.61 m/s^2:
a_total = sqrt((3.47)^2 + (9.61)^2),
a_total = sqrt(12.05 + 92.40),
a_total = sqrt(104.45),
a_total ≈ 10.22 m/s^2.
Therefore, the magnitude of the total acceleration of the car 8 seconds after the start is approximately 10.22 m/s^2.