First slide

Instantaneous acceleration

Question

The area under acceleration-time graph gives

Moderate

A

Distance travelled
B

Change in acceleration
C

Force acting
D

Change in velocity

Solution

dv/dt = a
$\large \therefore \int\limits_{v_1 }^{v_2 } {dv} = \int\limits_{t_1 }^{t_2 } {a\;dt}$

$\large \Rightarrow (v_2 - v_1 ) = \int\limits_{t_1 }^{t_2 } {a\;dt} = Area\;under\;a - t\;graph$

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