(a) From Newton's Second law of motion we know that
#"Force"="mass"xx"acceleration"#
Solving for acceleration Inserting given values we get
#veca=vecF/m#
#=>veca=1/10(9hati+12hatj)#
#=>|veca|=sqrt([9/10]^2+[12/10]^2)#
#=>|veca|=1/10sqrt(81+144)#
#=>|veca|=1/10sqrt225#
#=>|veca|=1.5ms^-2#
(b) (i) Using the kinematic expression
#vecs(t)=vec(s_0)+vecut+1/2vecat^2#
#vecs(5)=0+(11/5hat i+hatj)xx5+1/2xx1/10(9hati+12hatj)xx5^2#
#=>vecs(5)=11hat i+5hatj+45/4hati+15hatj#
#=>vecs(5)=(11+45/4)hat i+20hatj#
#=>vecs(5)=89/4hat i+20hatj#
#=>|vecs(5)|=sqrt((89/4)^2+(20)^2)#
#=>|vecs(5)|=sqrt(7921/16+400)#
#=>|vecs(5)|=sqrt(14321)/4m#
(ii) Using the kinematic equation
#vecv(t)=vecu+vec at#
Inserting given values we get
#vecv(t)=(11/5hat i+hatj) +1/10(9hati+12hatj)t#
(iii) North east direction is defined by a unit vector as
#(hati+hatj)#
General expression for velocity is
#vecv(t)=(11/5hat i+hatj) +1/10(9hati+12hatj)t#
To meet the condition we get at time #t#
#vecv(t) =n(hati+hatj)#
where #n# is a positive number. Comparing two expressions for #vecv(t)# we get
#11/5+9/10t=n#
#=>10n-9t=22# ......(1)
also
#1+6/5t=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#
