The three equations of motion are central to Class 9 and 11 physics and JEE mechanics. Enter the initial velocity, acceleration and time to get the final velocity and displacement, with the third equation used as a check.
Enter the initial velocity and acceleration.
Enter the time.
Read the final velocity and displacement.
Velocity: v = u + at
Displacement: s = ut + ½at²
Without time: v² = u² + 2as
The kinematics calculator solves motion in a straight line when the acceleration stays constant. You give it the starting velocity u, the acceleration a and the time t, and it returns the final velocity v, the displacement s and the average velocity over that interval. It also checks the answer against the third equation of motion, so you can see that all three relations agree.
These three relations, often called the SUVAT equations after the letters s, u, v, a and t, are taught in Class 9 and revisited in Class 11 before JEE and NEET mechanics. They answer everyday questions too: how far a scooter travels while braking, how fast a train is moving after pulling out of a station, or how long a ball thrown upwards keeps rising. Once you know any three of the five quantities, the other two follow.
1. Choose a positive direction, usually the direction of the starting motion, and give u, a and later v and s signs that match it.
2. Convert all values to SI units: velocity in m/s (divide km/h by 3.6), acceleration in m/s² and time in seconds.
3. Find the final velocity with v = u + at.
4. Find the displacement with s = ut + ½at².
5. Find the average velocity as s ÷ t, which for constant acceleration also equals (u + v) ÷ 2.
6. Check your work with v² = u² + 2as. Both sides should match, allowing for rounding.
Acceleration is the rate of change of velocity. If it is constant, velocity grows by the same amount every second, so after t seconds v = u + at. On a velocity-time graph this is a straight line, and displacement is the area under it: a rectangle of height u and width t, plus a triangle of base t and height at. That area is ut + ½at². Eliminating t between the two equations gives v² = u² + 2as, which is useful when time is not given.
The calculator returns displacement, the change in position with a sign. If a body slows down, stops and then moves backwards within the time you enter, s tells you only where it ends up relative to the start, not how far it travelled in total. A ball thrown up at 15 m/s with a = −9.8 m/s² for 3 s has a displacement of 0.9 m, but it rose about 11.5 m and fell about 10.6 m. For total path length, split the motion at the moment v = 0.
All three equations assume the acceleration is the same throughout the interval. Real cars accelerate hard at first and gently later, and falling objects are slowed by air resistance, so the equations give an idealised answer. For motion with changing acceleration you need calculus, v = ∫a dt and s = ∫v dt, or you break the journey into pieces that each have roughly constant acceleration. Circular motion also needs a different treatment, since the direction of velocity keeps changing.
Arjun rides his scooter at 12 m/s towards a red signal and brakes with a steady deceleration of 2.5 m/s² for 4 seconds.
First equation: v = u + at = 12 + -2.5 × 4 = 2 m/s
Second equation: s = ut + ½at² = 12 × 4 + 0.5 × -2.5 × 4² = 28 m
Check with the third equation: v² = u² + 2as → 4 = 4
Answer: Final velocity (v) 2 m/s; Displacement (s) 28 m; Average velocity 7 m/s
Mixing km/h with m/s. A speed of 54 km/h must be entered as 15 m/s.
Entering a positive acceleration for braking or for a ball thrown upwards, when it should be negative in the chosen sign convention.
Using the equations over a time that goes past the moment the object stops, for example a braking car, which does not reverse after stopping.
Treating the displacement as the total distance travelled when the motion changes direction.
Applying the equations to motion where acceleration clearly varies, such as a rocket burning fuel.
Solving Class 9 and Class 11 numericals on equations of motion.
Estimating braking distances for road safety lessons and driving theory.
Working out the height and flight time of a ball thrown straight up.
Checking lab results from a trolley on an inclined track with a ticker timer or motion sensor.
Rough planning of train or lift motion where acceleration is held nearly constant.
How do I handle a body thrown upwards?
Take upwards as positive and use a = −9.8 m/s².
Do these work if acceleration changes?
No, they assume constant acceleration.