A speedboat increases its speed from 50 ms-1 to 80 ms-1 in a distance of 200 m.
What is the the magnitude of its acceleration?
the time it takes the boat to travel this distance?
i am totally lost on my physics homework..
can some scientist help me out.. and show me the formulas /steps used to solve? thx
2007-11-03 23:50:32 · 3 answers · asked by BoBomakeup 2 in Science & Mathematics ➔ Physics
hmm... I think i get it ..
there're 2 more question that I am confused with.. :
A bullet of mass 0.015 kg leaves the barrel of a rifle with a speed of 800 ms-1 in 0.05 seconds.
Find the acceleration of the bullet as moves through the barrel of the rifle.
Find the force required to achieve this rate of acceleration.
and
A 20 kg block is initially at rest on a smooth horizontal surface. A horizontal force of 75 N is required to set the block in motion. Find the blocks rate of acceleration.
i am so frustrated.. becuase our teacher just threw the questions at us, and we didn't learn this part. now the hwk is due tomorrow!
2007-11-04 00:26:01 · update #1
change that question into feet miles per hour and i'll figure it up for you. I don't even know what a ms is. meters per second.???
that freakin flying. you must mean something else that would be like mach 6. 10,599 miles per hours. wow thats fast. what the heck kind of gokart are we gonna buld next summer :). I cant keep the chain on my 13 HP.
2007-11-04 00:06:43 · answer #1 · answered by hifi1863 2 · 0⤊ 1⤋
Let's assume uniform acceleration. Now: the average speed is 65 mps. Distance = Velocity x Time...Time = Distance / Vel.
Time = 200m/65mps; T = 3.08 seconds. (That's the second half of the question.)
It accelerated 30 mps in 3.08 seconds. Since acceleration is the change in velocity per unit of time the acceleration is 30mps/3.08s = 9.75 m/s^2
Okay?
2007-11-04 07:07:37 · answer #2 · answered by Richard S 6 · 0⤊ 0⤋
v = vo + at
where v = final velocity
vo = initial velocity
a = acceleration
t = time
at = v -vo ------eqn.1
at = 80 - 50 = 30 m/s
consider, s = vot + (1/2)at^2
s = (vo + at/2)t
t = s/(vo + at/2)
t = 200 /(50 + 30/2) = 3.07 sec. -----this is the time it takes to reach this distance.
substitute to eqn.1
at = 30
a = 30/3.07 = 9.75 m/s^2 -----this is the acceleration magnitude.
2007-11-04 07:11:28 · answer #3 · answered by dongskie mcmelenccx 3 · 0⤊ 0⤋