High School Physics
 

The driver of a car going 88.0 km/h suddenly sees the lights of a barrier 38.0 m ahead. It takes the driver 0.75 s to apply the brakes, and the average acceleration during braking is -10.0 m/s2.

He will hit the wall.

What is the maximum speed at which the car could be moving and not hit the barrier 38.0 m ahead? Assume that the acceleration doesn't change.

threve

...what?

57.

No...

Yes

The website rejected the answer. So, I guess not.

I hate WebAssign :(

wait does he hit the brakes with 38.0m to go or is 38.0m then he hits the brakes?

When he sees the barrier he is 38m away, then .75 seconds later he actually applies the brakes.

- Justin

Ok try 58.

ok thought soo

Not that either... :(

59?

...are you just guessing?

idk tsar...I came up with 59.4km/h

Well the answers are being beamed down to me from the Space Station via the satellite, so the numbers come in a little bit distorted.

Just trying to help.

w00t

at least it didn't come in as 75 :lol:

No its okay, its just that I have a limited number of tries...

How many more do you have left?

59...

No, you lie.

wait i think i got it... the maximum speed he can be going and stop in time is 19.84 m/s or 71.4 km/h

try that i misunderstood the first time lol

using

d = (v)(t) + (1/2)(a)(t)^2


thats how i came up with my number . . .

v = 88 km/h = 24.4 m/s reaction = 0.75s distance before brakes = d = (24.4 m/s)(0.75)s d = 18.3m to hit brakes total 38.0m - 18.3m leaves 19.7m to stop

t = sqrt (d)(2)/(a) t = sqrt (19.7)(2)/(10) t = 1.98s to stop in time

maximum Velocity he can go using d = (v)(t) + (1/2)(a)(t)^2

19.7 = (v)(1.98s)+(0.5)(-10m/s^2)(1.98s)^2
19.7 = 1.98v - 19.602 39.302 = 1.98v v = 19.84 m/s

That's not correct.

how so?