An object moving with uniform acceleration has a velocity of 10.5 cm/s in the po

Question

An object moving with uniform acceleration has a velocity of 10.5 cm/s in the positive x-direction when its x-coordinate is 2.72 cm. If its x-coordinate 2.30 s later is ?5.00 cm, what is its acceleration? The object has moved to a particular coordinate in the positive x-direction with a certain velocity and constant acceleration; then it reverses its direction and moves in the negative x-direction to a particular x-coordinate in time t. We are given an initial velocity vi = 10.5 cm/s in the positive x-direction when the initial position is xi = 2.72 cm (t = 0). We are given that at t = 2.30 s, the final position is xf = ?5.00 cm. The acceleration is uniform so that we have the following equation in terms of the constant acceleration a. xf ? xi=Vi*t-1/2at^2 Now we substitute the given values into this equation. (___cm)-(___cm)=(___cm/s)(____s)+1/2a(___s)^2

Explanation / Answer

Please change the values At time t = 0 x0 = 3cm v = 12 cm/s At time t=2s x2 = -5cm ?x = x2 - x0 = -8 cm ?t = 2 s For constant acceleration, ?x = v?t + ½ a(?t)² a = 2(?x - v?t) / (?t)² = 2(-8 - 24)/4 = -16 cm/s²

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