Question

If tangential acceleration vector of a speeding up particle
at P is 3î +4j cap m/s2 find the angle theta

(1) 53°
(2) 37°
(3) 30°
(4) None of these​

Answer 1

Answer:

tangential acceleration vector of a speeding up particle

at P is 3î +4j cap m/s2 find the angle theta

(3) 30°

Answer 2

Answer:

Angle made by acceleration vector is 53°

Explanation:

tangential acceleration of a speeding up particle  is given as $a=3\hat i+4\hat j$.

The x-component of acceleration is =3 $m/s^2$

the y-component of acceleration is= 4 $m/s^2$

tan(θ)= y component of a vector / x component of a vector

⇒$tan(\theta)=\frac{4}{3}$

⇒$\theta= tan^{-1}(\frac{4}{3} )=53.1$

The angel theta is approximately 53°. option (1) is correct.