Angle between velocity and acceleration vectors in the following cases are given below. Match the correct pairs.
List I List II
a) Vertically projected body
e) 900
b) For freely dropped body
f) changes from point to point
c) For projectile
g) zero
d) In uniform circular motion
h) 1800
a→h,b→g,c→f,d→e.
a→f,b→g,c→h,d→e.
a→e,b→f,c→h,d→g.
a→g,b→h,c→e,d→f.
A
a→h,b→g,c→f,d→e.
B
a→f,b→g,c→h,d→e.
C
a→g,b→h,c→e,d→f.
D
a→e,b→f,c→h,d→g.
Open in App
Solution
Verified by Toppr
(a) The direction of motion gives velocity which is in upward direction and the acceleration due to gravity acting in downward direction always hence angle is 1800.
(b) Body is moving in downward direction hence velocity is in downward direction and acceleration is acting downwards hence angle is zero.
(c) In projectile motion, particle has velocity changing from point to point but acceleration acting in downward direction always.
(d) In uniform circular motion velocity is always tangential to the circular path at each point and centripetal acceleration acting always inwards towards center hence angle is 900.
Was this answer helpful?
1
Similar Questions
Q1
Angle between velocity and acceleration vectors in the following cases are given below. Match the correct pairs.
List I List II
a) Vertically projected body
e) 900
b) For freely dropped body
f) changes from point to point
c) For projectile
g) zero
d) In uniform circular motion
h) 1800
View Solution
Q2
A man of mass m stands on a weighing machine in a lift List-IList-IILift movesreading of machinea) up with uniform acceleration 'a'e) mgb) down with uniform acceleration 'a'f) m(g+a)c) down with uniform velocityg) m(g−a)d) freely falling lifth) 0
View Solution
Q3
Match the items in the second list with corresponding items from the first list
If a:b = c:d, and e:f = g:h, then (ae+bf):(ae-bf)=?
View Solution
Q5
Match the classes of monosaccharides listed in column I with examples given in column II and select the right number showing the correct combination of alphabets of the two columns