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# Formulas Motion in One Dimension with Solved Examples

## Examples based on these formulas

### Questions based on v = u + at

### Answer

### Answer

### Question based on s = ut + 1/2 at^{2}

### Answer

### Question based on v^{2} = u^{2} + 2as

### Answer

**By one dimension we mean that the body is moving only in one plane and in a straight line.**

Like if we roll a marble on a flat table, and if we roll it in a straight line (not easy!), then it would be undergoing one-dimensional motion.

There are four variables which put together in an equation can describe this motion. These are Initial Velocity (u); Final Velocity (v), Acceleration (a), Distance Traveled (s) and Time elapsed (t). The equations which tell us the relationship between these variables are as given below.

v = u + at

v^{2} = u^{2} + 2as

s = ut + 1/2 at^{2}

average velocity = (v + u)/2

Of course, these equations are applicable only if acceleration is constant

Armed with these equations you can do wonderful things like calculating a cars acceleration from zero to whatever in 60 seconds !!

A cyclist speeds up at a constant rate from rest to 8 m/s in 6s. Find the acceleration of the cyclist.

u = 0 m/s [body starts from rest]

v = 8 m/s

t = 6 s

v = u + at

=> a = (v-u)/t = (8-0)/6 = 4/3 m/s^{2} = 1.33 m/s^{2}

An electron has a constant acceleration of 3.2 m/s^{2}. At a certain instant its velocity is 9.6 m/s.
What is its velocity 2.5 seconds earlier?

a = 3.2 m/s^{2}

v = 9.6 m/s

u = ?

t = 2.5 s

v = u + at

=> u = v - at = 9.6 -3.2 x 2.5 = 1.6 m/s^{2}

A racing car reaches a speed of 50 m/s. At this instant it begins a uniform negative acceleration using a parachute and a braking system and comes to rest 50 s later. How far does the car travel after acceleration starts.

u = 50 m/s

v = 0 m/s

t = 50 s

Here we have two unknowns 's' and 'a'. So first we will find 'a' using v = u + at.

0 = 50 + a x 5

=> 0 = 50 + 5a

=> a = -50/5 = -10 m/s^{2}

Now, we find the distance travelled

S = ut + 1/2 at^{2}

= 50 x 5 + 1/2 x (-10) x 5^{2})

= 250 -125 = 125 m

A car brakes from a speed of 108 km/h to 72 km/h duing a displacement of 100m. What is its acceleration?

We write down the given data and make the units consistent. All the units should be in SI

u = 108 km/h = 108 x 1000/(60 x 60) = 108 x 5/18 = 30 m/s

[1 km = 1000 m and 1 h = 60 x 60 = 3600 s]

v = 72 km/h = 72 x 5/18 = 20 m/s

s = 100 m

v^{2} - u^{2} = 2as

=> a = (v^{2} - u^{2})/(2 x 100) = (400 - 900)/(2 x 100) = -2.5 m/s^{2}