Question

Starting from rest, a boulder rolls down a hill with constant acceleration and travels 3.00m during the first second.

Answer 1

Answer:

a) 9.00 m b) 6.00 m/s  c) 12.00 m/s

Explanation:

a) If the acceleration is constant, and we know that the displacement during the first second was 3.00 m, as the boulder (assumed that we can treat it as a point mass) started from rest, we can say the following:

Δx = $\frac{1}{2}*a*t^{2}$ = 3.00 m

As t = 1 s, replacing in the expression above, and solving for a, we have:

$a = \frac{2*3.00m}{1s2}$ = 6.00 m/s²

In order to know how far it travels during the second second, we need to know the value of the speed after the first second, as it is the initial velocity when the second second begins:

vf = v₀ + a*t ⇒ vf = 0 + 6 m/s²*1s = 6.00 m/s

The total displacement, during the second second, will be as follows:

Δx = v₀*t + $\frac{1}{2}*a*t^{2}$ = 6.00m/s*1s +$\frac{1}{2}*6.00 m/s2*1s^{2} = 9.00 m$

⇒ Δx = 9.00 m

b) At the end of the first second, the final speed can be obtained as follows:

vf = v₀ + a*t ⇒ vf = 0 + 6 m/s²*1s = 6.00 m/s

c) At the end of the second second, the final speed can be obtained as follows:

vf = v₀ + a*t ⇒ vf = 0 + 6 m/s²*2s = 12.00 m/s