a body moving with uniform acceleration has a velocity of 15.6cm/s when its x coordinate is 4.28cm. if its x coordinate 2.82s later is -8cm, what is the magnitude of acceleration? answer in units of cm/s^2.
2007-09-18 19:53:03 · 4 answers · asked by michael v 1 in Science & Mathematics ➔ Physics
Under uniform acceleration:
x(t) = x0 + v0t + 1/2at^2
Solve for a:
a = 2(x(t) - x0 - v0t) / t^2
Plugnchug:
x(@ t=2.82 s) = -8 cm
x0 = 4.28 cm
v0 = 15.6 cm/s
2007-09-18 20:56:15 · answer #1 · answered by Anonymous · 0⤊ 0⤋
For constant acceleration,
s = s0 + v0t + (1/2)at^2
a = 2(s - s0 - v0t) / t^2
a = 2(- 8 - 4.28 - (15.6)(2.82) / (2.82)^2
a = 2(- 56.272) / 7.9524
a = -112.544 / 7.9524
a ≈ - 14.152 cm/s^2
2007-09-18 20:58:17 · answer #2 · answered by Helmut 7 · 0⤊ 0⤋
comparable 2.31 m/s^2 ?? if i think of two bodies accelerating in the direction of one yet another via gravity , they approch one yet another at a particular acc. besides , as quickly as the girl starts shifting she can not get purchase on the floor for to any extent further tug of conflict. ( that's an icy frictionless floor )
2016-11-05 21:28:29 · answer #3 · answered by ? 4 · 0⤊ 0⤋
Initial Velocity = 15.6 cm/sec
Final Velocity = -8 cm/sec
Dist.covered= (4.28 - 2.82)
.........................= 1.46 cm
v² - u² = 2as
(-8)² - (15.6)² = 2 * a * 1.46
64 - 243.36 = 2.92 a
- 179.36 = 2.92 a
a = (-179.36) / 2.92 = - 61.42 cm/Sec²
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2007-09-18 21:07:05 · answer #4 · answered by Joymash 6 · 0⤊ 0⤋