Ave/Instantateous Velocity

This is really an algebra problem.
First write what you have to show...
[tex]\displaystyle\frac{x_{10}-x_{0}}{t_{10}-t_{0}} \stackrel{?}{=} v_{5}[/tex]
Use the hint: "the formulae for constant acceleration". (My hint: use one of these formulas on the left for [tex]x_{10}[/tex], and use the other on the right.)

 

It's fine. Since the acceleration is constant, the instantaneous velocity has a simple form... which you will use on the right-hand side. By doing another integral, you'll get an expression for the instantaneous position that you'll use on the left-hand side. Suggestion don't use the specific value of 8m/s^2 now... stick to the algebra and keep it as "a".

 

If an object starts from rest and moves with a constant acceln of 8 m/s², how far does it travel in 10s?

 

Nope.
Sorry, but that is the final velocity.

One of the eqns of motion is,

vf = vi + at

where vf is the final velocity
vi is the initial velocity
a is the accln
t is the time of travel

Since we're starting from rest, then vi = 0. The accln = 8 m/s² and t = 10s, so
vf = 0 + 8*10
vf = 80 m/s
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Getting back to the problem though, you have to find the distance travelled in 10s. You should use the formula,

[tex]x_f = x_i + v_i*t + \frac{1}{2}at^2[/tex]

[tex]\mbox{Here,}\ x_i = 0\ \mbox{and}\ v_i = 0,\ \mbox{so the formula reduces to,}[/tex]

[tex]x_f = \frac{1}{2}at^2[/tex]

Putting in a = 8 and t = 10, you will get

[tex]x_f = 400m[/tex]
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[tex]\mbox{But}\ x_f\ \mbox{is the distance travelled at }t = 10[/tex]
[tex]\mbox{i.e. }x_f = x_{10}[/tex]
[tex]x_0 = 0[/tex]
[tex]t_{10} = 10[/tex]
[tex]t_0 = 0[/tex]
[tex]v_{av} = \frac{(x_{10} - x_0)}{(t_{10} - t_0)}[/tex]
[tex]v_{av} = \frac{(400 - 0)}{(10 - 0)}[/tex]
[tex]v_{av} = 40 m/s[/tex]

[tex]\mbox{Now you have to use one of the eqns of motion to show that the instantaneous velocity at }t = 5,\ v_5,\mbox{ has the same value as }v_{av}[/tex]