In common language, accelerate means to speed up. But in Physics, it has a very specific meaning. Acceleration is described as the rate of change of velocity of an object, irrespective of whether it speeds up or slows down. If it speeds up, acceleration is taken as positive and if it slows down, the acceleration is negative. It is caused by the net unbalanced force acting on the object, as per Newtonâ€™s Second Law. Acceleration is a vector quantity as it describes the time rate of change of velocity, which is a vector quantity. Acceleration is denoted by a. Its SI unit is m/s^{2} and dimensions are M^{0}L^{1}T^{â€“2}.
If v_{0}, v_{t} and t represents the initial velocity, final velocity and the time taken for the change in velocity, then, the acceleration is given by:
\[\vec a = \frac{{{{\vec v}_t}--{{\vec v}_0}}}{t}\]
In one dimensional motion, we can use; \[a = \frac{{{v_t}--{v_0}}}{t}\]
General formula
If \[\vec r\]represents displacement vector and \[\vec v = \frac{{d\vec r}}{{dt}}\]represents the velocity, then; acceleration:\[\vec a = \frac{{d\vec v}}{{dt}} = \frac{{{d^2}\vec r}}{{d{t^2}}}\]
In one dimensional motion, where x is the displacement, and \[v = \frac{{dx}}{{dt}}\]is the velocity, then; \[a = \frac{{dv}}{{dt}} = \frac{{{d^2}x}}{{d{t^2}}}\]
Example 1:
A car starts from rest and achieves a speed of 54 km/h in 3 seconds. Find its acceleration: Solution:
v_{0} = 0, v_{t} = 54 km/h = 15 m/s, t = 3s, a = ?
A body moves along the x- axis according to the relation x = 1 â€“ 2 t + 3t^{2}, where x is in meters and t is in seconds. Find the acceleration of the body when t = 3 s Solution:
We have:\[x = 1--2t + 3{t^2}\] Then; velocity\[v = \frac{{dx}}{{dt}} = --2 + 6t\] Acceleration:\[a = \frac{{dv}}{{dt}} = 6\,\]= 6 m/s^{2}. (We see that the acceleration is a constant here. Therefore, at t = 3s also, its value is 6 m/s^{2}.
Practice Question:
A car accelerates from rest at a constant rate $\alpha $for some time, after which it slows down at a constant rate $\beta $and comes to rest. If the total time elapsed is t, then the maximum velocity acquired by the car is (a) $\left( {\frac{{{\alpha ^2} + {\beta ^2}}}{{\alpha \beta }}} \right)\,t$ (b) $\left( {\frac{{{\alpha ^2} - {\beta ^2}}}{{\alpha \beta }}} \right)\,t$ (c)$\frac{{(\alpha + \beta )\,t}}{{\alpha \beta }}$ (d) $\frac{{\alpha \beta \,t}}{{\alpha + \beta }}$