Question

In the non-uniform circular motion shown below, the magnitude of change in tangential acceleration of the particle is

• A. $4m/s^2$
• B. $16m/s^2$
• C. $0m/s^2$
• D. $2m/s^2$

Answer 1

The correct option is A $4m/s^2$

Change in acceleration vector from A to B
$\Delta \overrightarrow{a}=\overrightarrow{a_B}-\overrightarrow{a_A}$
$\Rightarrow |\Delta \overrightarrow{a}|=\sqrt{a_t^2+a_t^2-2a_t{a}_t\cos \theta }$
$=\sqrt{2^2+2^2-2(2{)}^2\cos \pi }=\sqrt{8+8}=\sqrt{16}=4m/s^2$