(a) From Newton's Second law of motion we know that
Force=mass×acceleration
Solving for acceleration Inserting given values we get
→a=→Fm
⇒→a=110(9ˆi+12ˆj)
⇒∣∣→a∣∣=√[910]2+[1210]2
⇒∣∣→a∣∣=110√81+144
⇒∣∣→a∣∣=110√225
⇒∣∣→a∣∣=1.5ms−2
(b) (i) Using the kinematic expression
→s(t)=→s0+→ut+12→at2
→s(5)=0+(115ˆi+ˆj)×5+12×110(9ˆi+12ˆj)×52
⇒→s(5)=11ˆi+5ˆj+454ˆi+15ˆj
⇒→s(5)=(11+454)ˆi+20ˆj
⇒→s(5)=894ˆi+20ˆj
⇒∣∣→s(5)∣∣=√(894)2+(20)2
⇒∣∣→s(5)∣∣=√792116+400
⇒∣∣→s(5)∣∣=√143214m
(ii) Using the kinematic equation
→v(t)=→u+→at
Inserting given values we get
→v(t)=(115ˆi+ˆj)+110(9ˆi+12ˆj)t
(iii) North east direction is defined by a unit vector as
(ˆi+ˆj)
General expression for velocity is
→v(t)=(115ˆi+ˆj)+110(9ˆi+12ˆj)t
To meet the condition we get at time t
→v(t)=n(ˆi+ˆj)
where n is a positive number. Comparing two expressions for →v(t) we get
115+910t=n
⇒10n−9t=22 ......(1)
also
1+65t=n
⇒5n−6t=5 ......(2)
To solve (1) and (2), multiply (2) with 2 and subtract from (1)
⇒10n−12t=10 .....(3)
3t=22−10=12
t=4s
