Physics question??????????

Question

A truck covers 35.0 m in 9.00 s while smoothly slowing down to final speed 2.20 m/s.

A)find its original speed

B)find its accelaration

can someone please explain. i dont understand what is original speed

i was thinking in dividing 35 over 9 and then subtracting 2.20. please explain

Answer 1

distance equals the sum of inital and final velocity divided by 2 then multiplied by time

d=   v inital + v final

      _____________  X  T

               2

35=  vi  + 2.2

        _______      X9

             2                  

70=  vi + 2.2 X 9

7.7=  vi + 2.2

5.5 = vi

2.2-5.5

______

9

-0.37 m/sec squared'

First you have to find the inital speed use the distance formula and just use algebra

then find the acceleration, it is negitivein this case due to the fact this is a decell problem

Answer 2

s=35m,t=9s,V=2.2m/su=?m/s

V=u-at  0=2.2-a*9   a=  2.2/9  a=-0.244m/s^2

  V^2=u^2-2aS =     0^2=u^2-2*(-0.244*35)  =  u^2=8.556  u=2.93m/s

original speed or  initial velocity=2.93m/s

2)acceleration=-0.244 but since that is deceleration a=+0.244m/s^2

Answer 3

Don't just 'play' with the given numbers.  The idea is to figure out what general principle of physics applies to the situation first, and then put in the details.

"Smoothly slowing down" is a clue to me that they want us to assume the acceleration of the truck is constant.  If this is the case then we can apply the following equations from constant acceleration kinematics;

v = v0 + a t

x = x0 + v0 t + 1/2a t^2

I'll choose my position origin to make x0 = 0.  v0 is the original speed of the truck we are asked to find.  a is the acceleration we want to find.

At time t = 9s we know that v must be 2.20 m/s and x must be 35.0 m.

Plugging in the specifics,

2.20 m/s = v0 + a (9 s)

35.0 m = 0 + v0 (9 s) + 1/2a (9 s)^2

These two equations can be solved simultaneously for v0 and a.  I use QuickMath to do the solving for me.  It gives

v0 = 5.7778 m/s

a = - 0.375309 m/s^2 (negative because the acceleration is opposite the direction of the velocity, which I took to be positive.)