Answer:
No, the deviation in the path of asteroid is not by 22°
Step-by-step explanation:
Given:
velocity of asteroid,
![v_a=21\ km.s^(-1)](https://img.qammunity.org/2021/formulas/physics/college/ttnrt4k13ud5r3ebj132h7850zjwmwdbjl.png)
acceleration of the rocket,
![a=0.035\ km.s^(-2)](https://img.qammunity.org/2021/formulas/physics/college/4x6zmaxjf1wv2fszazjy7jdym5bmaycjcw.png)
time of acceleration,
![t=40\ s](https://img.qammunity.org/2021/formulas/physics/college/y2jk6ugbswcmmv66mtpdg4wuxondz1pl2p.png)
Now, the final velocity of the asteroid:
using the equation of motion,
![v=u+a.t](https://img.qammunity.org/2021/formulas/physics/high-school/xq18olz3cqeglc3cuukbd5ihjqc8p855wo.png)
where:
final velocity
initial velocity in the direction
![v=0+0.035* 40](https://img.qammunity.org/2021/formulas/physics/college/gxz59wukhqyw9ysuohn8n6r2xh40d5me5e.png)
![v=1.4\ km.s^(-1)](https://img.qammunity.org/2021/formulas/physics/college/s75dbvzyi32501xgsjv4xix9wcmuconpii.png)
Now direction of the resultant velocity:
![\tan\beta=(v_a)/(v)](https://img.qammunity.org/2021/formulas/physics/college/tblh0elpfqitsqct3rh14ka3625w9r690y.png)
![\tan\beta=(21)/(1.4)](https://img.qammunity.org/2021/formulas/physics/college/6sl75a0wc7ix2cziehe4jvem2bggh0e83p.png)
![\beta=86.186^(\circ)](https://img.qammunity.org/2021/formulas/physics/college/3v0nzvm5kq98fc8fmv9hgr60dm6bequtsh.png)
So, the deviation in the asteroid:
![\theta=90-\beta](https://img.qammunity.org/2021/formulas/physics/college/ixr77txn9bc2zvkc9yxtm5b3h2kui7mhkm.png)
![\theta=90-86.186](https://img.qammunity.org/2021/formulas/physics/college/chuey8xewz9hrhmp9aje4jjtbloidsfi5q.png)