A world-class sprinter accelerates to his maximum speed in 4.2s . He then maintains this speed for the remainder of a 100-m race, finishing with a total time of 9.3s .
What is the runner's average acceleration during the first 4.2 ?
What is his average acceleration during the last 5.1 ?
What is his average acceleration for the entire race?
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An alert hiker sees a boulder fall from the top of a distant cliff and notes that it takes 1.30s for the boulder to fall the last third of the way to the ground. You may ignore air resistance.
What is the height of the cliff in meters?
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A marble is released from one rim of a hemispherical bowl of diameter 55.0cm and rolls down and up to the opposite rim in 10.0s .
Find the average speed.
Find the average velocity of the marble.
Let's see your attempts first...
An alert hiker sees a boulder fall from the top of a distant cliff and notes that it takes 1.30s for the boulder to fall the last third of the way to the ground. You may ignore air resistance.
What is the height of the cliff in meters?
yf=yi + vi*t + .5*a*t^2
yf=final position
yi=initial position (this will represent the last third of the cliff)
vi=initial velocity (assumed to be 0)
t=time
a=acceleration (gravity, -9.8)
0=yi + 0*1.3 + (-4.6)*1.3^2
yi=7.774m
So the cliff is 3*yi = 23.322m
but its wrong
What is the height of the cliff in meters?
yf=yi + vi*t + .5*a*t^2
yf=final position
yi=initial position (this will represent the last third of the cliff)
vi=initial velocity (assumed to be 0)
t=time
a=acceleration (gravity, -9.8)
0=yi + 0*1.3 + (-4.6)*1.3^2
yi=7.774m
So the cliff is 3*yi = 23.322m
but its wrong
It's a 'free fall in Newtonian mechanics' then.
But you would still need to know what height that mountain was situated from the sea level.
As the acceleration due to gravity is 9.81 m/s squared, but that's near the surface of the earth.
But you would still need to know what height that mountain was situated from the sea level.
As the acceleration due to gravity is 9.81 m/s squared, but that's near the surface of the earth.
Your mistake is in assuming that the initial velocity is 0. After it has fallen 2/3 of the way, its velocity will certainly not be 0.
Gravity certainly can be approximated to 9.8ms^2, no problem with that.
I can't see a simple way to approach this problem, so my attempt might seem a bit complicated, but this is what I would do:
Gravity certainly can be approximated to 9.8ms^2, no problem with that.
I can't see a simple way to approach this problem, so my attempt might seem a bit complicated, but this is what I would do:
CODE
x = height of cliff
u = 0 (velocity at top of cliff)
v1 = (velocity at 2/3 down)
a = g
s1 = 2x/3 (untimed fall)
s2 = x/3 (1.30s fall)
t = 1.30s
u = 0 (velocity at top of cliff)
v1 = (velocity at 2/3 down)
a = g
s1 = 2x/3 (untimed fall)
s2 = x/3 (1.30s fall)
t = 1.30s
Using these variables, I set up:
CODE
(v1)^2 = (u^2) + 2a(s1), for the first part of the fall (1)
s2 = v1 + 0.5at^2, for the second part of the fall (2)
s2 = v1 + 0.5at^2, for the second part of the fall (2)
I re-wrote s1 and s2 in terms of x and made v1 the subject of eq.(1)
I then substituted for v1 in eq.(2) and it looks like I've got a quadratic in x after a bit of re-arranging.
That's a bit convoluted but the only way I can see! Does that approach match the level of teaching you've received so far or do you think the question should have a simpler solution?
