Physics question.

[JazzyJ]

The driver of a car going 90 km/h suddenly sees the lights of a barrier 40 m ahead. It takes the driver 0.75s tp apply the brakes, and the avg acceleration during braking is -10.o m/s^2.

a. determine whether the car hits the barrier

b. What is the maximum speed at which the car could be moving and not hit the barrier 40m ahead? Assume that the acceleration rate doesnt change

id really appriciate some help

 

[JazzyJ]

bumpity bump

 

[RenegadeZ3]

ahhhhh kinematic equations, but he's gone 30 m before he even applies the brakes, so he definitely hits the barrier, part 2 to come in a second

 

[JazzyJ]

for the second part i think you have to use

Vf^2=Vo^2+2a (delta T)

 

[RenegadeZ3]

ok at 90 km/h and all that junk for part a) he goes 80 meters, so he hits the barrier

b)the max speed he can go is 28.28 m/s

if you want to know how, i can explain it pretty

 

[RenegadeZ3]

shit my last post got cut short, but i was going to say that i can explain it pretty easily if you want, but that is the formula you use

 

[*BeckyK*]

Why does your teacher like to mix up KM and M. I hate it when teachers do that. It sucks. Kbye

 

[JazzyJ]

i would love if you could explain it, If possible. Really i just need the equation

 

[SteezePatrol]

no he stops exactly 5 meters before

the speed he needs is 93.5 km to hit it

also, i just pulled that outta my ass

 

[DunDunDun]

okay, i like just did this in my class. yo in grade 11? haha

you gotta use like number 3 i think? i dunno. but its not that hard. jsut use an example outta the book.

 

[JazzyJ]

im in the 11th grade. There arent any examples like this in my book

 

[zepscratch]

too bad i dont take A.P physics until next year

 

[Chubby.Chaser]

Vfinal^2 = Vinitial^2 + 2a?d

when it stops Vfinal = 0

you need to make sure velocity and acceleration are using same units for distance and time i.e. a = n * m/s^2 and

v = n * m/s not km/h

then add the distance travelled before starting the acceleration

0.75s * Vinitial in m/s not km/h

 

[pmills]

be happy, i'm almost positive this is the first homework question to be answered before going to the next page

 

[Anathema]

ha, i'm in engineering (well....was) this shit is my life. but i'm lazy. apologies.

 

[mk2]

i think its 31.25

 

[Badfish2]

90km/h*.75s = 18.75 m

40-18.75m = 31.25m -- distance that the brakes are applied till barrier.

a = -10m/s

vi = 90km/h => 25m/s

vf = 0

solving for the distance it takes to stop:

d = vf^2-vi^2

--------

2a

d = 31.25m

distance to barrier - distance to stop = 0

so, he'll stop just in time before crashing into the barrier, but will just touch the barrier.

 

[Badfish2]

wait, my bad. he'll wreck into the barrier. 40-18.75 = 21.25

distance it would take him to stop is 31.25m but he only has 21.25 m of space. so if he isn't wearing a seatbelt he's fucked.