course PHY 121
9/9
A ball starting from rest rolls 11 cm down an incline on which its acceleration is 22 cm/s2, then onto a second incline 49 cm long on which its acceleration is constant, and on which it spends 1.561 seconds. How much time does it spend on the first incline and what is its acceleration on the second?vvvv
vf^2 = v0^2 + 2 a `ds. Vf^2 = (0)+ 2(22cm/s^2)(11cm) = 484cm^2/s^2 – sqrt --- = 22m/s vF. good
A = `dv/`dt, so 22m/s^2 = (22m/s)/`dt. [22m/s^2]/[22m/s] = 1s.
22 m/s is the average velocity, not the change in velocity, so unless the change in velocity happens to equal your average velocity (which it doesn't in this case, and usually won't) your equation won't give you an accurate value of `dt
[22m/s^2]/[22m/s] = 1 s^-1, not 1 s, and this calculation therefore doesn't give you units of time.
You didn't solve the equation correctly for `dt. The correct solution would have been
`dt = [22m/s]/[22m/s^2] = 1s.
Still, this is based on an incorrect value for the change in velocity so it isn't the correct `dt for this motion.
`dt = 1.561s. `ds = 49cm. v = `ds/`dt = 49cm/1.561 = 31.4cm/s. a = `dv/`dt = (31.4cm/s)/(1.561s) = 20 cm/s^2.
Your attempt at direct reasoning is pretty good, but it doesn't hold up to close scrutiny and doesn't give you a correct result.
The main problem is that you are not distinguishing average velocity from instantaneous velocity or change in velocity, which are three different quantities.
The word 'velocity' must always be modified. The symbol v stands for instantaneous velocity.
`ds / `dt is not instantaneous velocity, but average velocity.
Neither the instantanous velocity nor the average velocity is the same thing as `dv, which is the change in velocity.
To solve this problem you need to use the equations of motion. The initial velocity on this second ramp is the same as the final velocity on the first. You know the acceleration and the length of the ramp. So you know v0, a and `ds. Using the equations of motion, you can find vf and `dt.
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