Calculating speed from acceleration and distance?

Question

A 2.67 kg block slides down a frictionless plane, from rest, with an acceleration of 5.48009 m/s^2. What's the block's speed after travelling 2.1 m along the incline? Answer in m/s

I tried using unit analysis to figure it out and got
v=(distance x acceration)^1/2 = 3.392372319 m/s but the computer says its wrong?

I don't know if I am even on the right track. Please help!

Awesome, thanks so much guys!

Answer 1

V^2=Vo^2+2a(X-Xo)

X-Xo=d
Vo=0 (the block started from rest

V^2=2ad

V=sqrt(2ad)

Answer 2

Vf = Final Velocity Vi = Initial Velocity a = Acceleration d = Distance Vf^2 = Vi^2 + 2ad Plug in the values 0 = Vi^2 + 2(3.9)(290) Solve for Vi: Vi = sqroot (2262) Vi = 47.56 m/s

Answer 3

Well we need to solve for time to start with

For this we use the following formula

x(t) = xo + vot + 0.5at^2

Let xo and vo both be equal to zero

So then we get

2.1 m = 0.5*(5.48009 m/s^2) t^2

Solving for t yields

t = 0.8754488966 s

Now we go to our next formula (which is the derivative of the first one we used for constant acceleration)

v(t) = vo + at

v(t) = (5.48009 m/s^2)(0.875 s)

v = 4.797538744 m/s

Rounds to app,roximately

v = 4.80 m/s

Answer 4

2.1=.5*5.48009*t^2
t=0.87545

v(t)=5.48009*0.87545

=4.7975 m/s

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