A man is in a car moving with an acceleration of `5m*s^(-2)` . Find the apparent value of the acceleration due to gravity and the direction of pull of the earth with respect to him.

Explanation

Let the acceleration of the man in the car be `veca` and acceleration due to gravity be `vecg`. Hence acceleration due to gravity relation to the man is, `vecg. =vecg-veca=vecg+(vec(-a))`. From Fig.2.44,

`g.=sqrt(g^2+a^2)=sqrt(9.8^2+5^2)`
`=sqrt(121.04)=11m*s^(-2)`
The angle `theta` , which `vecg.` makes with `vecg` is given by
`tan theta=a/g=5/(9.8)=0.51=tan27^@`
`therefore theta =27^@`
Hence, with respect to the man the earth.s pull is acting downward at an angle `27^@` with the vertical in the side opposite to his direction of motion.