Initial velocity of object vi=10.0 cm/s
initial position fo vector of the object is xi=3.09 cm
Final position of vector xf=-5.00cm
then displacement of object s = xf-xi=-5.00-3.09=-8.09cm
time t=2.55 s
s=vit+1/2at2
-5.00 = 11*2.55+1/2*a2.552
a = (-5.00 - 10*2.55*2)/2.552 = 2.94 cm
Acceleration is 2.94 cm.
<h3>What is a
cceleration?</h3>
Speed increase is the name we provide for any cycle where the speed changes. Since speed is a speed and a bearing, there are simply two different ways for you to speed up: change your speed or shift your course or change both. In mechanics, speed increase is the pace of progress of the speed of an item concerning time. Speed increases are vector amounts. The direction of an item's speed increase is given by the direction of the net power following up on that article. An item's typical speed increase throughout some stretch of time is its adjustment of speed separated by the term of the period. Numerically, quick speed increase, in the meantime, is the constraint of the typical speed increase over a little time period. In the terms of analytics, immediate speed increase is the subordinate of the speed vector concerning time.
Learn more about acceleration, refer:
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Answer:
Whether the force exerted by the locomotive on the wall was larger
Than the force the locomotive could exert on the wall.
Explanation:
The Newton's third law of motion States that every force have it's equal and opposite reaction force, whose magnitude is the same as the applied force. Therefore the magnitude of these opposite forces will be equal.
So we have;
F12=-F21
F12 is the force in a direction
-F21 is the force in the opposite direction.
Therefore we see that the magnitude of the force the locomotive exerts on the wall is equal to the one the wall exerts on the locomotive. Both magnitudes are equal but in opposite directions.
M = mass of the girl = 53.4 kg
m = mass of skateboard = 3.55 kg
V' = velocity of the combination of girl and skateboard before she jumps forward = 0 
V = velocity of the girl forward = 1.32
v = velocity of the skateboard afterward = ?
Using conservation of momentum
(M + m) V' = MV + m v
inserting the values
(53.4 + 3.55) (0) = (53.4) (1.32) + (3.55) v
v = - 19.86
