Any stratagies??

Question

A 1000kg car racing up a mountain road runs out of gas at a height of 35m while traveling at 24m/s. The driver shifts into neutral and coasts onward. Neglecting all frictional losses.

1.Will he clear the 65m summit?


2. His brakes fail; at what speed will he reach the bottom of whichever side of the mountain he rolls down?

Answer 1

mgh2-mgh1=mv2^2 - mv1^2
g(h2-h1)=v2^2 - v1^2

1)
9.8(h2-h1)=24^2-0
h2-h1 = 58.8 > 65-35 then answer is yes

2)
9.8(35-0)=V^2 - 24^2
V=30.3

Answer 2

Since we can neglect all frictions, then the speed can 100% be applied to overcome gravity. So, this 24 m/s speed could, using a ramp, be converted to pure vertical motion, with only gravity slowing down the car.
The equations that applies are:
vf=v0 + at
s=v0 t + 1/2 a t^2
where v0 is the initial velocity, a is the decelerative effect of gravity (9.81 m/s^2) and t is the time.

We want vf (final speed) to be zero, so
0 = 24 - 9.81 t
(we put the acceleration negative as in this case, it is slowing the car down)
24 = 9.81 t
or t=2.446 seconds

plugging this in the other equation

s = 24 * 2.446 - 1/2 9.81 2.446^2
s = 58.704 - 29.346 = 29.358 m

so the car should be able to climb 29.358 m above its current position. It is now at 35 m, can go an additional 29.358 m, for a total of 64.358 m... Mhhh; looks like the car will stall just as it reaches the top.

Now, falling back. Since the car should come back down just as fast as it went up, all we need to know is how much extra speed is gained from dropping 35 m. Or how much speed is gained from dropping from 64.358 m from zero speed.


s = 1/2 a t^2
64.358 = 1/2 9.81 t^2
this gives t= 3.622 seconds
and with
v= a t
v = 9.81 * 3.622 = 35.5 m/s



(ormoz, above, forgot that the kinetic energy is 1/2 m v^2; that is why his answer is off as he is missing the 1/2 factor, although his approch of equalling kinetic and potentail energy is just as valid)

Answer 3

For part 1, you know the initial velocity and the acceleration (due to gravity), so you can write-out the equation describing the velocity of the car between v=24 m/s and v=0 m/s (i.e., when the forward motion stops). I would find the time t at which the velocity equals zero, then integrate with respect to t over that interval to see if the distance is greater than 30 meters.

I think when you solve 1, 2 becomes fairly easy.