Question

a) Find the angle between the directions of the velocity and acceleration vectors
at time 4 of a particle with position vector r = (t^2 + 1)i { – 2tj +(t^2– 1)K -​

Answer 1

r
=(t
2
−4t+6)
i
^
+(t
2
)
j
^
​

velocity =
dt
dr
​
=(2t−4)
i
^
+(2t)
j
^
​

acceleration =
dt
dv
​
=(2)
i
^
+(2)
j
^
​

a
.
v
=((2t−4)
i
^
+(2t)
j
^
​
).((2)
i
^
+(2)
j
^
​
)
when they are perpendicular then a.v = 0
((2t−4)
i
^
+(2t)
j
^
​
).((2)
i
^
+(2)
j
^
​
)=0
(2t−4)×2+4t=0
4t−8+4t=0
t=1s
solution



HOPE IT HELPS