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.
UNKNOWN =
a decimal and a unit of measure
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