Both planets are similar in shape and have a rocky surface. Not sure about the phases though
At the point of maximum displacement (a), the elastic potential energy of the spring is maximum:

while the kinetic energy is zero, because at the maximum displacement the mass is stationary, so its velocity is zero:

And the total energy of the system is

Viceversa, when the mass reaches the equilibrium position, the elastic potential energy is zero because the displacement x is zero:

while the mass is moving at speed v, and therefore the kinetic energy is

And the total energy is

For the law of conservation of energy, the total energy must be conserved, therefore

. So we can write

that we can solve to find an expression for v:
Explanation:
It is given that,
The acceleration of the toboggan, 
Initial speed of the toboggan, u = 0
We need to find the distance covered by the toboggan. Using the second equation of motion as :

At t = 1 s


At t = 2 s


At t = 3 s


Hence, this is the required solution.
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(Example 1 )
<span>If the Voltage that furnishes the current is an ideal (no internal resistance) Voltage source. Then; </span>
<span>V/R = i </span>
<span>V/2R = i/2 If external resistance doubles, current reduced to 1/2 of original value </span>
<span>V/3R = i/3 If external resistance triples, current reduced to 1/3 of original value </span>
<span>(Example 2) </span>
<span>But if the Voltage that furnishes the current is a practical [contains an internal resistance (Ri)] Voltage source. Then the current is a function of the Voltage source`s internal resistance, which does not double nor triple, plus the external resistance which is being doubled and tripled. </span>
<span>V/(R + Ri) = i </span>
<span>V/(2R + Ri) = greater than i/2 but less than I. </span>
<span>V/(3R + Ri) = greater than i/3 but less than i/2</span>