A car traveling at a speed of vo = 42 m/s stops smoothly (that is, its deceleration is constant) over a distance of d = 176 m.

I got all of them right but I can't get the right answer for letter d.

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a) What is its acceleration during the time it is stopping? (Be careful about the sign!)

a = m/s2 *

-5 OK

HELP: Assume that the car's acceleration is constant.

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b) How long (what amount of time) does it take for the car to come to a stop?

tstop = s *

8.4 OK

HELP: Which equation relates velocities, acceleration, and time?

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c) After the car has gone 1/3 of the stopping distance, what is its speed?

v1/3 = m/s *

34.3 OK

HELP: Dx = d/3

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d) If the car had been going 10 m/s but had stopped in the same d = 176 m distance, what would its acceleration have been?

a(vo = 10m/s) = m/s2

-.57 NO

-.28 NO

HELP: Use same concept as in part (a).

Physics braking car. PLEASE HELP!?

The fundamental equations here for constant acceleration from rest are

v = a t

d = a t^2 /2

so we find d = v^2 /(2a).

t is the time since starting to move, v is the velocity, d is the distance from the point of rest.

Now the braking car is just the same situation except with time running backwards. So we can use the same equations except that some of the answers may have the wrong sign and you have to pay attention.

What we notice is that: for fixed d, a is proportional to v^2. So you can take the correct answer from Part a) and multiply it by (10/42)^2.

I get (-)0.284... m/s^2

So your 2nd answer should be RIGHT!

Check: time to brake is v/a = 35.2 s

distance travelled is a t^2/2 = 175.9... m - seems to work out.

Physics braking car. PLEASE HELP!?

you have:

v=v0+at, with v0=10m/s, and v=0 (it stops at the end), as such you have v0=-at, with negative 'a'.

next, the distance it travels is the integral of the speed:

d=int_t0^t1 (v0+at)dt= d0 + v0t+ at^2 |_t0^t1

d0=0, at=-v0, thus

d=2*v0t=176m=2*10*t, so t=176/20=88s, and so a=-v0/t=-0.11(36) m/s^2

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