Acceleration Calculator

What's the key thing to understand about The Five SUVAT Equations?

For uniform acceleration, five equations relate the five variables (s=distance, u=initial velocity, v=final velocity, a=acceleration, t=time). v = u + at. s = ut + ½at². s = vt − ½at². v² = u² + 2as. s = ½(u+v)t. Any two known values allow you to calculate the other three. Named after their variables, these equations are fundamental to classical mechanics and form the core of GCSE and A-level physics.

What should I know about Free Fall and g?

Near Earth's surface, objects in free fall (ignoring air resistance) accelerate at g = 9.81 m/s². Dropped from rest: after 1 second a speed of 9.81 m/s, after 2 seconds 19.62 m/s, falling 4.9 m in the first second and 19.6 m in the first two seconds. Terminal velocity occurs when air resistance equals gravitational force — approximately 53 m/s (190 km/h) for a typical human skydiver in the belly-to-earth position, higher in the head-down position.

What's the key thing to understand about Real-World Examples?

A car accelerating from 0 to 60 mph (26.8 m/s) in 6 seconds: a = 26.8/6 = 4.47 m/s². That is 0.46g (gravity equivalents). A fighter jet can sustain 9g briefly. An airliner at takeoff experiences approximately 0.3g. A passenger braking emergency stop (0.8g) is approximately 7.8 m/s². The Large Hadron Collider accelerates protons to 0.9999% the speed of light — the SUVAT equations no longer apply at relativistic speeds; Einstein's equations are needed.

What do I need to know about Limitations of SUVAT?

SUVAT equations assume constant (uniform) acceleration — they cannot model variable acceleration without calculus. Real-world situations like a car accelerating (changing gear, engine power curve) or a falling object (air resistance increasing with speed) require integration of the force equation over time. For many practical engineering and physics problems, however, assuming constant acceleration over short time intervals gives adequately accurate results.

What's the key thing to understand about Deriving the SUVAT Equations?

The five SUVAT equations aren't five independent facts to memorise separately — they all follow from just two basic definitions of motion with constant acceleration. The first building block is the definition of acceleration itself: a = (v − u) / t, rearranged to give v = u + at (final velocity equals initial velocity plus acceleration times time). The second building block is the definition of average velocity for constant acceleration: since velocity changes uniformly from u to v, the average velocity during the motion is simply (u + v)/2, and distance travelled is average velocity times time: s = ((u+v)/2) × t. Every other SUVAT equation can be derived by combining or substituting between just these two relationships — for example, substituting v = u + at into s = ((u+v)/2)t and simplifying algebraically produces s = ut + ½at², while eliminating time between the two building blocks produces v² = u² + 2as. Recognising that all five equations trace back to these two simple ideas makes it far easier to reconstruct any SUVAT equation from first principles, rather than relying purely on memorisation.

What's the difference between Acceleration and Deceleration?

Acceleration, in physics, refers to any change in velocity over time — not just speeding up. Deceleration (slowing down) is simply acceleration that acts in the opposite direction to the object's current motion, reducing its speed rather than increasing it. This is why a negative value for acceleration in the SUVAT equations doesn't automatically mean 'slowing down' — it only means the acceleration vector points in the negative direction, as defined by whichever direction was chosen as positive at the start of the problem. If an object is already moving in the negative direction and experiences a negative acceleration, it's actually speeding up (accelerating), not decelerating, because the acceleration reinforces rather than opposes the existing motion. This is why clearly defining a positive direction at the start of any SUVAT problem — and staying consistent with that choice throughout — is essential: a car braking while moving forward has a negative acceleration (deceleration) if forward is defined as positive, but the same braking car reversing would have a positive acceleration relative to the same sign convention, since it's still slowing its (negative) velocity toward zero.

What's the key thing to understand about Non-Uniform Acceleration?

The SUVAT equations only apply when acceleration is constant. When acceleration itself changes with time — a rocket burning fuel and losing mass, a car's acceleration tapering off as it nears top speed — velocity and position must instead be found from a = dv/dt and v = dx/dt using calculus: integrating a(t) once gives v(t), and integrating again gives position. Graphically, this is the same as saying the area under an acceleration-time graph gives the change in velocity, and the area under a velocity-time graph gives displacement, regardless of whether the graph is a straight line or a curve.

What's the difference between Average and Instantaneous Acceleration?

The SUVAT equations assume acceleration is constant, which makes 'the acceleration' during an interval unambiguous — it's the same throughout. When acceleration actually varies, it's important to distinguish average acceleration, Δv/Δt (the overall change in velocity divided by the total time), from instantaneous acceleration, a = dv/dt (the rate of change of velocity at one specific instant, found using calculus). A sprinter's instantaneous acceleration is largest in the first stride out of the blocks and tapers toward zero as they approach top speed, even though their average acceleration over the whole race is a single, much smaller, blended number. Velocity-time graphs make the distinction visual: average acceleration is the gradient of the straight line connecting the start and end points, while instantaneous acceleration is the gradient of the tangent to the curve at one particular moment — the two match only when the graph is already a straight line, i.e. when acceleration really is constant.

What should I know about Typical Magnitudes of Acceleration?

Converting between m/s² and 'g' (multiples of Earth's 9.81 m/s² gravitational acceleration) gives an intuitive feel for how extreme a given acceleration actually is:

Sustained accelerations above about 5g risk blood draining from a pilot's brain (G-LOC, or g-induced loss of consciousness) without training and a pressurised g-suit, which is why fighter pilots train specifically to tolerate the sustained high-g turns their aircraft are capable of.

What should I know about a Worked Example: A Three-Stage Journey?

Consider a train journey with three distinct phases. Phase 1: the train accelerates uniformly from rest to 25 m/s over 50 seconds. Phase 2: it cruises at a constant 25 m/s for 120 seconds. Phase 3: it decelerates uniformly to rest over 20 seconds. Phase 1 acceleration: a = (v−u)/t = (25−0)/50 = 0.5 m/s². Distance covered: s = ½(u+v)t = ½(0+25)×50 = 625m. Phase 2 distance (constant velocity, so a=0): s = vt = 25×120 = 3,000m. Phase 3 deceleration: a = (0−25)/20 = −1.25 m/s². Distance covered: s = ½(25+0)×20 = 250m. Total distance: 625 + 3,000 + 250 = 3,875m (3.875 km), covered in a total time of 50+120+20 = 190 seconds. The journey's average velocity across the whole trip is therefore 3875/190 ≈ 20.4 m/s — noticeably less than the 25 m/s cruising speed, because the slower acceleration and deceleration phases drag the average down. This kind of multi-stage calculation, applying a different constant acceleration value to each distinct phase of motion and adding the results, is exactly how real transport timetables, lift-off profiles, and vehicle performance figures are actually built up from the basic SUVAT equations.

What should I know about Centripetal Acceleration in Circular Motion?

Not all acceleration involves speeding up or slowing down — an object moving at constant speed around a curved path is still accelerating, because acceleration is any change in velocity, and velocity includes direction as well as speed. This centripetal acceleration points toward the centre of the curve and has magnitude a = v²/r, where v is speed and r is the radius of the path. For a car taking a roundabout at 10 m/s with a radius of 15m: a = 10²/15 = 100/15 ≈ 6.67 m/s² (about 0.68g) directed toward the roundabout's centre, which is why sharper bends taken at the same speed demand noticeably more grip from the tyres, and why racing drivers brake before a corner rather than during it — centripetal acceleration adds to whatever braking deceleration is also present, and a tyre's grip can only supply so much total acceleration in any direction before it slides.