Question

zwei Autos starten gleichzeitig von derselben stelle. Auto A beschleunigt mit a1=2m/s^(-2) und hat nacht Δt=15s einen Vorsprung von Δs=50m vor Auto B . Welche beschleunigung a2 hat Auto B ?

Answer 1

I dont speak german, but from google translate

"Two cars start at the same time by the same agency . A car accelerates with a1 = 2m / s ^ (- 2) and has adversely At = 15s a lead of .DELTA.s = 50m in front of car B . What acceleration a2 has car B"

is this more or less it?

Answer 2

Sorry . I will try to speak English..

the right is : ( car A has accelerates with a1=2m/s^-2 ) , not ( A car accelerates with a1 = 2m / s ^- 2 )

Answer 3

From the diagram, this really is the question required :)

can you help me to solve it , please ?

Answer 4

I speak a little german, give me a second

Answer 5

I think you meant ^(2) instead of ^(-2)
You can find the distance that car A has traveled in 15 seconds, then you subtract 50 from that to find the distance car B has traveled in the same 15 second. Assuming car B also has a constant acceleration, you can use the same equation to find car B's acceleration:
I got a2 = 14/9 m/(s^2)

Answer 6

it ^(-2) not ^(2) , in fact that is my problem with the quetion !

Answer 7

that's an error, they probably meant to write \(ms^{-2}\) which is \(m/s^2\)

Answer 8

at first i think that some thing error too , but when i asked my professor he tell me their is no error , the question is correct !

Answer 9

that can't be, those arent units for acceleration

Answer 10

meters per second per second = m / (s^2)
\[m/s^2 = m * s ^{-2}\]

Answer 11

i know that! but when the professor tell me their is no mistake , i doubt that the wrong from me :D

Thank you both very much :")

Answer 12

how can i find the distance that car A has traveled in 15 seconds when i don't know the
value of ( v ) ? * v=vi+aΔt

Answer 13

This is how i would do it,


use: \(\Delta x=v_0\Delta t+\dfrac{1}{2}a\Delta t^2\)

For car A: \(50~m+R=(0)(15~s)+\dfrac{1}{2}(2~m/s^2)(15~s)^2\)

\(R=\dfrac{1}{2}(2~m/s^2)(15~s)^2-50~m\)

For car B: \(R=(0)(15~s)+\dfrac{1}{2}a(15~s)^2\)

Sub in result from car A

\(\dfrac{1}{2}(2~m/s^2)(15~s)^2-50~m= \dfrac{1}{2}a(15~s)^2\)

\(a=\dfrac{\dfrac{1}{2}(2~m/s^2)(15~s)^2-50~m}{\dfrac{1}{2}(15~s)^2}=(2~m/s^2)-\dfrac{50~m}{\dfrac{1}{2}(15~s)^2}\)

Answer 14

Thank you very much :))

Answer 15

no problem!