randFromArray(['a', 'd', 't', 'v_i', 'v_f']) randFromArray(['a', 'd', 't', 'v_i', 'v_f']) randRangeNonZero(-200, 200) / 10 randRange(-400, 400) / 10 randRange(100, 200) / 10 roundTo(2, V_INIT * TIME + (1/2) * ACCEL * TIME * TIME) roundTo(2, V_INIT + ACCEL * TIME) /* negative under radical */ (OMITTED === 'v_i' && UNKNOWN === 't' && (V_FINAL * V_FINAL - 2 * ACCEL * DISP < 0.1)) || (OMITTED === 'v_f' && UNKNOWN === 't' && (V_INIT * V_INIT + 2 * ACCEL * DISP < 0.1)) || (OMITTED === 't' && UNKNOWN === 'v_i' && (V_FINAL * V_FINAL - 2 * ACCEL * DISP < 0.1)) || (OMITTED === 't' && UNKNOWN === 'v_f' && (V_INIT * V_INIT + 2 * ACCEL * DISP < 0.1)) || /* division by 0 */ (OMITTED === 'd' && UNKNOWN === 'a' && TIME === 0) || (OMITTED === 'd' && UNKNOWN === 't' && ACCEL === 0) || (OMITTED === 'v_i' && UNKNOWN === 'v_f' && TIME === 0) || (OMITTED === 'v_i' && UNKNOWN === 'a' && TIME === 0) || (OMITTED === 'v_i' && UNKNOWN === 't' && ACCEL === 0) || (OMITTED === 'v_f' && UNKNOWN === 'v_i' && TIME === 0) || (OMITTED === 'v_f' && UNKNOWN === 'a' && TIME === 0) || (OMITTED === 'v_f' && UNKNOWN === 't' && ACCEL === 0) || (OMITTED === 'a' && UNKNOWN === 'v_i' && TIME === 0) || (OMITTED === 'a' && UNKNOWN === 'v_f' && TIME === 0) || (OMITTED === 'a' && UNKNOWN === 't' && (-0.1 < V_INIT + V_FINAL) && (V_INIT + V_FINAL < 0.1)) || (OMITTED === 't' && UNKNOWN === 'd' && ACCEL === 0) || (OMITTED === 't' && UNKNOWN === 'a' && DISP === 0) (OMITTED === 't' && UNKNOWN === 'v_i') || (OMITTED === 't' && UNKNOWN === 'v_f') (OMITTED === 'v_i' && UNKNOWN === 't' && (V_FINAL * V_FINAL - 2 * ACCEL * DISP > 0.1)) || (OMITTED === 'v_f' && UNKNOWN === 't' && (V_INIT * V_INIT + 2 * ACCEL * DISP > 0.1))

d = DISP \text{m}

v_i = V_INIT \frac{\text{m}}{\text{s}}

v_f = V_FINAL \frac{\text{m}}{\text{s}}

a = ACCEL \frac{\text{m}}{\text{s}^2}

t = TIME \text{s}

OMITTED = {?}

UNKNOWN = {?}

Solve for UNKNOWN. Round to the nearest tenth.

Make sure you select the proper units. You may do arithmetic with a calculator. If there is more than one correct solution, you may enter either one.

roundTo(1, DISP) roundTo(1, V_INIT) roundTo(1, V_FINAL) roundTo(1, ACCEL) roundTo(1, TIME) m m/s m/s m/s² s
roundTo(1, -V_INIT) roundTo(1, -V_FINAL) m/s
roundTo(1, ((-V_FINAL) + sqrt(V_FINAL * V_FINAL - 2 * ACCEL * DISP))/(-ACCEL)) roundTo(1, ((-V_INIT) + sqrt(V_INIT * V_INIT + 2 * ACCEL * DISP))/ACCEL) s
roundTo(1, ((-V_FINAL) - sqrt(V_FINAL * V_FINAL - 2 * ACCEL * DISP))/(-ACCEL)) roundTo(1, ((-V_INIT) - sqrt(V_INIT * V_INIT + 2 * ACCEL * DISP))/ACCEL) s

UNKNOWN =

a decimal and a unit of measure

-9.8 rand(2) ? 0 : randRange(-100, 300) / 10 randRange(3, 200)/10 roundTo(2, V_INIT * TIME + (1/2) * ACCEL * TIME * TIME) roundTo(2, V_INIT + ACCEL * TIME)
0 randRange(5, 25) V_INIT randRange(1, 25) TIME * V_INIT false

v_f = v_i + at

v_f = V_INIT \frac{\text{m}}{\text{s}} + (ACCEL \frac{\text{m}}{\text{s}^2})(TIME \text{s})

v_f = roundTo(1, V_FINAL) \frac{\text{m}}{\text{s}}

v_f - at = v_i

V_FINAL \frac{\text{m}}{\text{s}} - (ACCEL \frac{\text{m}}{\text{s}^2})(TIME \text{s}) = v_i

roundTo(1, V_INIT) \frac{\text{m}}{\text{s}} = v_i

\dfrac{v_f - v_i}{t} = a

\dfrac{V_FINAL \frac{\text{m}}{\text{s}} - V_INIT \frac{\text{m}}{\text{s}}}{TIME \text{s}} = a

roundTo(1, ACCEL) \frac{\text{m}}{\text{s}^2} = a

\dfrac{v_f - v_i}{a} = t

\dfrac{V_FINAL \frac{\text{m}}{\text{s}} - V_INIT \frac{\text{m}}{\text{s}}}{ACCEL \frac{\text{m}}{\text{s}^2}} = t

roundTo(1, TIME) \text{s} = t

d = v_f t - \frac{1}{2}at^2

d = (V_FINAL \frac{\text{m}}{\text{s}})(TIME \text{s}) - \frac{1}{2}(ACCEL \frac{\text{m}}{\text{s}^2})(TIME \text{s})^2

d = roundTo(1, DISP) \text{m}

\dfrac{d + \frac{1}{2} at^2}{t} = v_f

\dfrac{DISP \text{m} + \frac{1}{2}(ACCEL \frac{\text{m}}{\text{s}^2})(TIME \text{s})^2}{TIME \text{s}} = v_f

roundTo(1, V_FINAL) \frac{\text{m}}{\text{s}} = v_f

\dfrac{d - v_f*t}{-\frac{1}{2}t^2} = a

\dfrac{DISP \text{m} - (V_FINAL \frac{\text{m}}{\text{s}})(TIME \text{s})}{-\frac{1}{2}(TIME \text{s})^2} = a

roundTo(1, ACCEL) \frac{\text{m}}{\text{s}^2} = a

0 = -\frac{1}{2}a*t^2 + v_f*t - d

By the quadratic formula:

t = \dfrac{ -v_f \pm \sqrt{ v_f^2 - 2ad } }{-a}

t = \dfrac{-(V_FINAL \frac{\text{m}}{\text{s}}) \pm \sqrt{(V_FINAL \frac{\text{m}}{\text{s}})^2 - 2(ACCEL \frac{\text{m}}{\text{s}^2})(DISP \text{m})}}{-(ACCEL \frac{\text{m}}{\text{s}^2})}

The quadratic has two solutions. Without making any assumptions about the direction of v_i, either one could be correct. Intuitively, you can imagine throwing an object upward or downward in such a way that they will both have the same downward velocity at the same point in space, but the one thrown downward will get there sooner.

t = roundTo(1, ((-V_FINAL) + sqrt(V_FINAL * V_FINAL - 2 * ACCEL * DISP))/(-ACCEL)) \text{s} \text{ or } roundTo(1, ((-V_FINAL) - sqrt(V_FINAL * V_FINAL - 2 * ACCEL * DISP))/(-ACCEL)) \text{s}

t = TIME \text{s}

d = v_i t + \frac{1}{2}at^2

d = (V_INIT \frac{\text{m}}{\text{s}})(TIME \text{s}) + \frac{1}{2}(ACCEL \frac{\text{m}}{\text{s}^2})(TIME \text{s})^2

d = roundTo(1, DISP) \text{m}

\dfrac{d - \frac{1}{2}at^2}{t} = v_i

\dfrac{DISP \text{m} - \frac{1}{2}(ACCEL \frac{\text{m}}{\text{s}^2})(TIME \text{s})^2}{TIME \text{s}} = v_i

roundTo(1, V_INIT) \frac{\text{m}}{\text{s}} = v_i

\dfrac{d - v_i t}{\frac{1}{2} t^2} = a

\dfrac{DISP \text{m} - (V_INIT \frac{\text{m}}{\text{s}})(TIME \text{s})}{\frac{1}{2}(TIME \text{s})^2} = a

roundTo(1, ACCEL) \frac{\text{m}}{\text{s}^2} = a

0 = \frac{1}{2} at^2 + v_i t - d

By the quadratic formula:

t = \dfrac{ -v_i \pm \sqrt{v_i^2 + 2ad} }{a}

t = \dfrac{-(V_INIT \frac{\text{m}}{\text{s}}) \pm \sqrt{(V_INIT \frac{\text{m}}{\text{s}})^2 + 2(ACCEL \frac{\text{m}}{\text{s}^2})(DISP \text{m})}}{ACCEL \frac{\text{m}}{\text{s}^2}}

The quadratic has two solutions. Without making any assumptions about the direction of v_f, either one could be correct. Intuitively, this is because an object moving upward at velocity v at one point in time will eventually be falling at the same speed (-v) at a different point in time.

t = roundTo(1, ((-V_INIT) + sqrt(V_INIT * V_INIT + 2 * ACCEL * DISP))/ACCEL) \text{s} \text{ or } roundTo(1, ((-V_INIT) - sqrt(V_INIT * V_INIT + 2 * ACCEL * DISP))/ACCEL) \text{s}

t = TIME \text{s}

d = \frac{1}{2}(v_i + v_f)t

d = \frac{1}{2}(V_INIT \frac{\text{m}}{\text{s}} + V_FINAL \frac{\text{m}}{\text{s}})(TIME \text{s})

d = roundTo(1, DISP) \text{m}

\dfrac{2d}{t} - v_f = v_i

\dfrac{2(DISP \text{m})}{TIME \text{s}} - V_FINAL \frac{\text{m}}{\text{s}} = v_i

roundTo(1, V_INIT) \frac{\text{m}}{\text{s}} = v_i

\dfrac{2d}{t} - v_i = v_f

\dfrac{2(DISP \text{m})}{TIME \text{s}} - V_INIT \frac{\text{m}}{\text{s}} = v_f

roundTo(1, V_FINAL) \frac{\text{m}}{\text{s}} = v_f

\dfrac{2d}{v_i + v_f} = t

\dfrac{2(DISP \text{m})}{(V_INIT \frac{\text{m}}{\text{s}}) + (V_FINAL \frac{\text{m}}{\text{s}})} = t

roundTo(1, TIME) \text{s} = t

v_f^2 = v_i^2 + 2ad

\dfrac{v_f^2 - v_i^2}{2a} = d

\dfrac{(V_FINAL \frac{\text{m}}{\text{s}} )^2 - (V_INIT \frac{\text{m}}{\text{s}} )^2}{2(ACCEL \text{s})} = d

roundTo(1, DISP) \text{m} = d

\pm\sqrt{v_f^2 - 2ad} = v_i

\pm\sqrt{(V_FINAL \frac{\text{m}}{\text{s}})^2 - 2(ACCEL \frac{\text{m}}{\text{s}^2})(DISP \text{m})} = v_i

Without making any assumptions about t, either direction for v_i could be correct. Imagine throwing an object upward at velocity v compared to throwing it downward at the same speed (-v). In both cases it will reach the same final velocity v_f at the same point in space; it will just take different amounts of time to get there.

v_i = roundTo(1, V_INIT) \frac{\text{m}}{\text{s}} \text{ or } roundTo(1, -V_INIT) \frac{\text{m}}{\text{s}}

However, we also know a = 0 \frac{\text{m}}{\text{s}^2}

We know intuitively that with zero acceleration v_i must equal v_f, so v_i = roundTo(1, V_INIT) \frac{\text{m}}{\text{s}}

v_f = \pm\sqrt{v_i^2 + 2ad}

v_f = \pm\sqrt{(V_INIT \frac{\text{m}}{\text{s}})^2 + 2(ACCEL \frac{\text{m}}{\text{s}^2})(DISP \text{m})}

Without making any assumptions about t, either direction for v_f could be correct. Imagine an object traveling upward at velocity v_i. It will reach a particular velocity v on the way up, then reach the same speed in the opposite direction (-v) on the way down at the same point in space; it will just take a different amount of time to get there.

v_f = roundTo(1, V_FINAL) \frac{\text{m}}{\text{s}} \text{ or } roundTo(1, -V_FINAL) \frac{\text{m}}{\text{s}}

However, we also know a = 0 \frac{\text{m}}{\text{s}^2}

We know intuitively that with zero acceleration v_f must equal v_i, so v_f = roundTo(1, V_FINAL) \frac{\text{m}}{\text{s}}

\dfrac{v_f^2 - v_i^2}{2d} = a

\dfrac{(V_FINAL \frac{\text{m}}{\text{s}})^2 - (V_INIT \frac{\text{m}}{\text{s}})^2}{2(DISP \text{m})} = a

roundTo(1, ACCEL) \frac{\text{m}}{\text{s}^2} = a