Acceleration Physics question?

Question

At time= 0 an automobile already traveling at 30km/hr. accelerates at 3m/s for 2 seconds. During there two seconds, how far does it go?

What is the formula to figure this out?

Answer 1

xf = x0 + v0t + 1/2 at^2

x0 is initial position (ignore it)

v0 is initial velocity

a is acceleration (danger--you've got wrong units on yours)

t is time

Plug and chug.

Answer 2

X=1/2 At^2 +Vo t + Xo

This equation says that the postion "X" of the object is equal to one-half of the acceleration (A) times the square of the time (t) plus the inital velocity (Vo) times the time (t again) plus the inital postition (Xo). If you stop and think middle term (intial velocity times time) makes intuitive sense as does the Xo term. The acceleration term is less intuitive and sometimes you just have to memorize that stuff until you see where its coming from.

The starting postion (Xo) is zero so we don't need that term.

You have to get the units consistant or the equation will not work. Most of it is in m and sec so we need to convert 30km/hr.

30 km/hr * (1000 m/km) * (hr/3600 sec)= 8.33 m/sec

I assume that the acceleration is 3m/s^2 as 3 m/s is not an acceleration.

X=1/2 At^2 + Vo t
X=.5(3 m/s^2) (2 sec)^2 + (8.33 m/sec)* 2 sec
X=.5*3*4 m + 16. 67 m
X= 6.0 m + 16.67 m = 22.67 m

you should check my arithmetic

enjoy

Answer 3

The formula is S=ut +(1/2)a(t^2)
where s = distance travelled
u = initial vel (in m/s)......30km/h=8.3333333m/s
t = time
a = acceleration

therefore distance = 8.333333 x 2 + 0.5 x 3 x (2 x 2)

=22.6666666 m

=22 and 2 thirds

Answer 4

Acceleration= Velocity(final)-Velocity(initial)
divided by
time

Answer 5

30km/hr => 8.33m/s

x=1/2at^2+vt
x= 1/2(3m/s^2)(2s)^2 + (8.33m/s)(2s)
x= 6m+16.66m
x= 22.66m