The speed of a point on a rotating turntable, which is 0.264 m from the center, changes at a constant rate from rest to 0.902 m/s in 1.94 s.
At t1 = 0.64 s, find the magnitude of the tangential acceleration.
Answer in units of m/s2
I assumed the velocity was constant so i divided .902/the time .64 to get acceleration at that time =1.41m/s^2 but this was incorrect...what am i doing wrong???
To find the magnitude of the tangential acceleration at t1 = 0.64 s, we need to calculate the rate of change of velocity over time.
Since the speed of the point on the rotating turntable changes at a constant rate, we assume that the acceleration is constant for the given time interval.
First, let's find the initial velocity at t1 = 0.64 s. We know that the speed changes at a constant rate from rest (0 m/s), so the initial velocity at t=0.64 s is given by:
v1 = v0 + a * t
where
v1 is the final velocity at t1 = 0.64 s,
v0 is the initial velocity at t0 = 0 s (which is 0 m/s for a rotating turntable),
a is the constant acceleration, and
t is the time interval, which is 0.64 s.
Applying the above equation:
v1 = 0 + a * 0.64
v1 = 0.64a m/s
Now, since the speed changes from rest to 0.902 m/s in 1.94 s, we can find the acceleration (a) using the equation:
v2 = v1 + a * t
where
v2 is the final velocity at t2 = 1.94 s, which is 0.902 m/s,
v1 is the initial velocity at t1 = 0.64 s, which is 0.64a m/s, and
t is the time interval, which is 1.94 - 0.64 = 1.3 s.
Applying the above equation:
0.902 = 0.64a + a * 1.3
To solve for a, we can rearrange the equation as:
1.3a + 0.64a = 0.902
Adding the like terms:
1.94a = 0.902
Dividing both sides by 1.94:
a = 0.902 / 1.94
a ≈ 0.465 m/s^2
Therefore, the magnitude of the tangential acceleration at t1 = 0.64 s is approximately 0.465 m/s^2.