Physics questions, just need help on a few things?

Question

The acceleration of a motorcycle is given by a(t) =1.46 t - 0.114t^2 . The motorcycle is at rest at the origin at time t=0.

(Now, i found the velocity equation as a function of time as well as the position, but its asking me this:)

Calculate the maximum velocity it attains.

Thanks guys for this one in advance.

Here's another one.

A large boulder is ejected vertically upward from a volcano with an initial speed of 39.7 . Air resistance may be ignored.

The boulder is moving upwards at a speed of 20.5 m/s @ time t=1.96 secs.

@ time t=6.14 secs, the boulder is moving @ a speed of 20.5 m/s downwards.

It is asking me When is the displacement of the boulder from its initial position zero?
Take the free fall acceleration to be = 9.80 .

And I keep arriving at the wrong answer. Any suggestions will help. Thanks a lot guys :D

i forgot to ask this but on the second question with the boulder, its asking to calculate the velocity of the boulder zero and it should be in seconds..i dont understand how this could be seeing velocity is in m/s. please help!

Answer 1

You get the velocity v(t) by integrating accel a(t). To find the max velocity, you differentiate v(t), which get you a(t). Max or min is obtained by making dv/dt =0, ie, a(t) =0

1.46t = 0.114 t *t, t=0 or t = 1.46/0.114, which is about 12.8 sec

Since the object is at rest at t=0, there is no velocity at t=0

From integration, v(t) = C + 0.73t*t - 0.038t*t*t and C=0
v(t) = t*t* (0.73 - 0.038*t)

You put t= 12.8 sec and find v(max) = 39.9

You find that as t increases beyond 12.8 sec, the quantity (0.73 - 0.038t) begin to drop fast

Answer 2

i'm not sure what the original question is asking. if it gives you a formula for acceleration then input how much time has passed and then plug that into v=at (v=v+at , but since you have 0 initial velocity just use v=at)... and you have your velocity. if the question doesn't give an elapsed time then plot a 3 dimensional graph ;)


part 2: ok, ignore all other velocities except the intial 39.7 (m/s i'm assuming)... again we have v=v+at ... what you want to use is your final velocity to be 0 and your initial velocity to be 39.7 ... acceleration is -9.8 ... so you have 0=39.7-9.8t ... solve for t... and you get approximately 4.051 ... this is to the high point when the object stops going up and then starts to go down... now, a nifty little tidbit is that when dealing with simple y axis movement and only the force of gravity it will take the object exactly the same amount of time to return to where it was initially launched from and when it reaches there it will be moving at the same velocity it was launched from. you can do the calculations using the distance formula d=vt + .5at^2 ... now that we know the time and then you set the initial velocity to zero in the same equation and use gravity for the acceleration and solve for t, and well... you get the same time and then using the velocity equation you get the same velocity (negating air resistance of course). so, your total trip time (or displacement zero - fancy wording) is 8.102 seconds.