Kinematics (SUVAT) calculator

Kinematics calculator for the SUVAT equations: enter three of displacement, initial and final velocity, acceleration and time to get the other two.

হালনাগাদ ২৮ সেপ্টেম্বর, ২০২৬যাচাই করা উদাহরণ: ৬

Final velocity v

m/s

সার্থক অঙ্ক: ৬; নিকটতম; সমান দূরত্বে শূন্য থেকে দূরে

Displacement s: 100m
Initial velocity u: 0m/s
Acceleration a: 2m/s²
Time t: 10s

Displacement is 100 m and final velocity is 20 m/s. The shaded area under the velocity line equals the displacement.

Velocity against time

Displacement against time

যেভাবে হিসাব করা হয় S

Knowns in SI units
$u = 0\ \mathrm{m/s},\quad a = 2\ \mathrm{m/s^2},\quad t = 10\ \mathrm{s}$
Final velocity
$v = u + at = 0 + (2)(10) = 20\ \mathrm{m/s}$
Displacement
$s = ut + \tfrac12 a t^2 = (0)(10) + \tfrac12(2)(10)^2 = 100\ \mathrm{m}$

Kinematics (SUVAT) calculator সম্পর্কে

Motion with constant acceleration is described by five quantities: displacement s, initial velocity u, final velocity v, acceleration a and time t. Each of the five SUVAT equations leaves one of them out, so any three known values fix the other two. When time is unknown, s = ut + ½at² is a quadratic and can have two valid roots.

Physics students use it for braking, free-fall and launch problems. With the defaults, an object starting from rest and accelerating at 2 m/s² for 10 s reaches 20 m/s and covers 100 m. A ball thrown straight up at 20 m/s passes 15 m twice, at 0.99 s and 3.09 s, and both times are listed.

Acceleration must stay constant and the motion must lie along one line. Values are signed: choose a positive direction and give opposing quantities a minus sign, so free fall with up as positive uses a = −9.80665 m/s².

সমাধান করা উদাহরণ

From rest at 3 m/s² for 8 s

Find: s and v (know u, a, t)
Initial velocity u: 0 m/s
Acceleration a: 3 m/s²
Time t: 8 s

Final velocity v: 24 m/s
Displacement s: 96 m

যাচাইয়ের উৎস: Python 3.8 decimal: v = 0 + 3·8 = 24; s = ½·3·8² = 96

Ball thrown up at 20 m/s passes 15 m twice

Find: v and t (know s, u, a)
Displacement s: 15 m
Initial velocity u: 20 m/s
Acceleration a: -9.81 m/s²

Time t: 0.990719 s
Second time (other root): 3.08675 s
Final velocity v: 10.2811 m/s
Final velocity at the second time: -10.2811 m/s

যাচাইয়ের উৎস: Python 3.8 decimal: t = (20 ∓ √(400 − 2·9.81·15))/9.81 = 0.9907186…, 3.0867534…; v = 20 − 9.81t = ±10.2810505…

Car braking from 10 m/s to rest in 50 m

Find: a and t (know s, u, v)
Displacement s: 50 m
Initial velocity u: 10 m/s
Final velocity v: 0 m/s

Time t: 10 s
Acceleration a: -1 m/s²

যাচাইয়ের উৎস: Python 3.8 decimal: t = 2s/(u+v) = 10; a = (0 − 100)/(2·50) = −1

Zero acceleration (uniform motion)

Find: v and t (know s, u, a)
Displacement s: 100 m
Initial velocity u: 5 m/s
Acceleration a: 0 m/s²

Time t: 20 s
Final velocity v: 5 m/s

যাচাইয়ের উৎস: Python 3.8 decimal: t = s/u = 20 (linear case of ½at² + ut − s = 0)

প্রশ্ন

What does SUVAT stand for?

SUVAT names the five quantities in the constant-acceleration equations: s for displacement, u for initial velocity, v for final velocity, a for acceleration and t for time. The equations are v = u + at, s = ut + ½at², v² = u² + 2as, s = ½(u + v)t and s = vt − ½at². Each omits a different variable, so you pick the one that leaves out the quantity you neither know nor need.

When can you use the SUVAT equations?

Only when acceleration is constant and the motion is in a straight line. Free fall near the Earth's surface qualifies while air resistance is small; standard gravity is 9.80665 m/s², the value adopted by the 3rd General Conference on Weights and Measures in 1901. A skydiver nearing terminal velocity, or a car whose acceleration fades as it gains speed, needs calculus or a numerical model instead.

How do you calculate stopping distance with SUVAT?

Set the final velocity to zero in v² = u² + 2as, which gives s = u² ÷ (2 × deceleration). A car braking from 10 m/s (36 km/h) at 1 m/s² stops in 50 m after 10 s. Doubling the starting speed quadruples the braking distance. The distance covered during the driver's reaction time comes on top and is uniform motion, s = ut.

Why are there two answers for time?

When time is the unknown, s = ut + ½at² is a quadratic in t and can have two positive roots. A ball thrown upward at 20 m/s with a = −9.81 m/s² is 15 m above the start at 0.99 s on the way up and again at 3.09 s on the way down, moving at 10.28 m/s each time but in opposite directions. Negative roots are dropped because time starts at zero.

What value of g should I use?

Use 9.80665 m/s², the defined standard gravity, unless the problem states another value; many textbooks round it to 9.81 or 9.8 m/s². Real sea-level gravity varies with latitude, from about 9.780 m/s² at the equator to 9.832 m/s² at the poles in the WGS 84 model. That spread changes answers by about 0.5%.

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নির্ভুলতা আপনার ইনপুট ও পদ্ধতির অনুমানের ওপর নির্ভর করে। দশমিক গণনায় 50টি সার্থক অঙ্ক ব্যবহৃত হয়, কিন্তু আনুমানিক হিসাব, সংখ্যাগত পদ্ধতি ও উৎসের তথ্য কম নির্ভুল হতে পারে; প্রদর্শিত মান রাউন্ড করলে এই সীমাবদ্ধতাগুলি দূর হয় না। স্বতন্ত্র উৎসের সমাধানের সঙ্গে যাচাই করা উদাহরণ: ৬। যেমন, “From rest at 3 m/s² for 8 s” উদাহরণটি Python 3.8 decimal: v = 0 + 3·8 = 24; s = ½·3·8² = 96-এর সঙ্গে যাচাই করা হয়।

এই পদ্ধতির উৎস কী?

OpenStax University Physics Volume 1, §3.4 Motion with constant acceleration; HyperPhysics — Motion equations for constant acceleration.

এই ক্যালকুলেটর সম্পর্কে

v = u + at,\quad s = ut + \tfrac12at^2,\quad v^2 = u^2 + 2as,\quad s = \tfrac12(u+v)t,\quad s = vt - \tfrac12at^2

উৎস

তথ্যসূত্রের সঙ্গে যাচাই করা

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