# Car Acceleration

If you know the initial and final velocity of a car (or whatever) - and the time used - the average acceleration can be calculated as

a = dv / dt

= (v_{ f }- v_{ s }) / dt (1)

where

a = acceleration of object (m/s^{ 2 }, ft/s^{ 2 })

dv = change in velocity (m/s, ft/s)

v_{ f }= final speed (m/s, ft/s)

v_{ s }= start speed (m/s, ft/s)

dt = time used (s)

Common benchmark velocities for acceleration of cars and motorcycles are

*0 - 60 mph = 0 - 26.8 m/s = 0 - 96.6 km/h**0 - 100 km/h = 0 - 27.8 m/s = 0 - 62.1 mph*

### Online Car Acceleration Calculator

** km/h **

** Note ** that force, work and power are calculated for mass acceleration only. Forces due to air resistance ( drag ) and rolling friction are not included.

#### mph

### Car Acceleration Diagram - * km/h *

Download and Print Car Acceleration Chart

### Car Acceleration Diagram - * mph *

Download and Print Car Acceleration Chart

If you know the distance moved and the time used - the acceleration can be calculated as

a = 2 ds / dt^{ 2 }(2)

where

ds = distance moved (m, ft)

### Acceleration of some known cars

### Acceleration Force

The acceleration force can be calculated as

* F = m a (3) *

* where *

* F = acceleration force (N, lb _{ f } ) *

* m = mass of car (kg, slugs) *

### Acceleration Work

The acceleration work can be calculated as

* W = F l (4) *

* where *

* W = work done (Nm, J, ft lb _{ f } ) *

* l = distance moved (m, ft) *

### Acceleration Power

The acceleration power can be calculated as

* P = W / dt (5) *

* where *

* P = power (J/s, W, ft lb _{ f } /s) *

### Example - Car Acceleration

A car with mass * 1000 kg (2205 lb _{ m } ) * accelerates from

*0 m/s (0 ft/s)*to

*27.8 m/s (100 km/h, 91.1 ft/s, 62.1 mph)*in

*10 s*.

The acceleration can be calculated from eq. 1 as

* a = ((27.8 m/s) - (0 m/s)) / (10 s) *

* = 2.78 m/s ^{ 2 } *

The acceleration force can be calculated from eq. 3 as

* F = (1000 kg) (2.78 m/s ^{ 2 } ) *

* = 2780 N *

The distance moved can be calculated by rearranging eq. 2 to

* ds = a dt ^{ 2 } / 2 *

* = (2.78 m/s ^{ 2 } ) (10 s) ^{ 2 } / 2 *

* = 139 m *

The acceleration work can be calculated from eq. 4 as

* W = (2780 N) (139 m) *

* = 386420 J *

The acceleration power can be calculated from eq. 5 as

* P = (386420 J) / (10 s) *

* = 38642 W *

* = 38.6 kW *

The calculation can also be done in ** Imperial units ** :

The acceleration can be calculated from eq. 1 as

* a = ((91.1 ft/s) - (0 ft/s)) / (10 s) *

* = 9.11 ft/s ^{ 2 } *

In the Imperial system mass is measured in slugs where * 1 slug = 32.17405 lb _{ m } *

The acceleration force can be calculated from eq. 3 as

* F = (( 2205 lb _{ m } ) (1/32.17405 (slugs/ lb _{ m } )) ) (9.11 ft/s ^{ 2 } ) *

* = 624 lb _{ f } *

The distance moved can be calculated by rearranging eq. 2 to

* ds = a dt ^{ 2 } / 2 *

* = (9.11 ft/s ^{ 2 } ) (10 s) ^{ 2 } / 2 *

* = 455 ft *

The acceleration work can be calculated from eq. 4 as

* W = (624 lb _{ f } ) (455 ft) *

* = 284075 ft lb _{ f } *

*1 ft lb*_{ f }= 1.36 J

The acceleration power can be calculated from eq. 5 as

* P = (284075 ft lb _{ f } ) / (10 s) *

* = 28407 ft lb _{ f } /s *

*1 ft lb*_{ f }/s = 1.36 W = 0.00182 hp

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