At what time is the velocity perpendicular to the acceleration

In summary, the velocity and acceleration vectors for a mass moving with constant acceleration in the x-y plane can be represented by Vx(t) = 3+t, Vy(t) = -4+t, and ax = ay = 1. The time at which the velocity is perpendicular to the acceleration is t=0.5 sec.
  • #1
hhhp8cec1
7
2

Homework Statement

:[/B]
A mass is moving with constant acceleration in the x-y plane. At time t = 0s the mass’s position is 2mˆı+3mjˆ, the mass’s velocity is 3 m/s ˆı −4 m / s jˆ and the mass is subject to an acceleration of 1 m/ s ^2 ˆı + 1 m/ s^ 2 i jˆ.

At what time is the velocity perpendicular to the acceleration?

Homework Equations

:
[/B]
v = v0 + α t
x = x0 + v0t + 0.5at^2
v^2= v0^2 +2a(x-x0)

The Attempt at a Solution

:
[/B]
I don't really know how to approach this question. I tried to take the magnitude of velocity and acceleration and find the radius using a= v^2/R, but I don't think that is the way to solve it.

I know that the answer is t=0.5 sec.

Many thanks for your help

 
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  • #2
In terms of the time t, what is the x component of velocity at time t? In terms of the time t, what is the y component of velocity at time t? In terms of the time t, what is the velocity vector at time t? If two vectors are perpendicular, what can we say about their dot product?
 
  • #3
hhhp8cec1 said:

Homework Statement

:
A mass is moving with constant acceleration in the x-y plane. At time t = 0s the mass’s position is 2mˆı+3mjˆ, the mass’s velocity is 3 m/s ˆı −4 m / s jˆ and the mass is subject to an acceleration of 1 m/ s ^2 ˆı + 1 m/ s^ 2 i jˆ.
[/B]

Homework Equations

:
[/B]
v = v0 + α t
x = x0 + v0t + 0.5at^2
v^2= v0^2 +2a(x-x0)

The Attempt at a Solution

:

I don't really know how to approach this question. I tried to take the magnitude of velocity and acceleration and find the radius using a= v^2/R, but I don't think that is the way to solve it.

I know that the answer is t=0.5 sec.

Many thanks for your help
[/B]
In vectors, what is the mass's velocity at time t?
 
  • #4
V(t)x= 3+t
V(t)y= -4+t
V = √(3+t)^2+(t-4)^2 , is that correct ?
A = √2
dot product: V ⋅ A = 0 because the angle is 90 deg.
But When I am trying to solve it I get an unsolvable quadratic equation...
 
  • #5
hhhp8cec1 said:
V(t)x= 3+t
V(t)y= -4+t
V = √(3+t)^2+(t-4)^2 , is that correct ?
A = √2
dot product: V ⋅ A = 0 because the angle is 90 deg.
But When I am trying to solve it I get an unsolvable quadratic equation...
You need to work with vectors. What is the vector ##\vec{v(t)}##? What is the acceleration vector ##\vec{a}##?
 
  • #6
I think I solved it, but I am not sure I did it correctly:

Vx(t) = 3+t
a x = 1

Vy(t) = -4+t
ay=1

cross product:
(3+t)⋅1 + (t-4)⋅1=0
t=0.5 sec
 
  • #7
hhhp8cec1 said:
I think I solved it, but I am not sure I did it correctly:

Vx(t) = 3+t
a x = 1

Vy(t) = -4+t
ay=1

cross product:
(3+t)⋅1 + (t-4)⋅1=0
t=0.5 sec
Right, but you mean dot product.
 
  • #8
hhhp8cec1 said:
I think I solved it, but I am not sure I did it correctly:

Vx(t) = 3+t
a x = 1

Vy(t) = -4+t
ay=1

cross product:
(3+t)⋅1 + (t-4)⋅1=0
t=0.5 sec
Yes, that's correct (note: you mean the *dot* product, not the cross product).
 

Related to At what time is the velocity perpendicular to the acceleration

1. What is the difference between velocity and acceleration?

Velocity refers to the rate of change of an object's position with respect to time, while acceleration refers to the rate of change of an object's velocity with respect to time.

2. How does velocity become perpendicular to acceleration?

When an object is moving in a circular path, at any point on the circle, the velocity will be tangential to the circle, while the acceleration will point towards the center of the circle. This creates a perpendicular relationship between the two.

3. Can velocity and acceleration be perpendicular at any other time besides during circular motion?

No, for velocity and acceleration to be perpendicular, the object must be moving in a circular path. In any other type of motion, the two vectors will have some degree of parallel or non-perpendicular relationship.

4. How does the magnitude of velocity and acceleration affect their perpendicular relationship?

The magnitude of velocity does not affect its perpendicular relationship with acceleration. However, the greater the magnitude of acceleration, the tighter the circular path will be, and the faster the object will be moving, resulting in a higher velocity.

5. How is the perpendicular relationship between velocity and acceleration useful in physics?

The perpendicular relationship between velocity and acceleration allows for the calculation of centripetal force, which is required for an object to maintain circular motion. It also allows for the prediction of an object's path and speed in circular motion.

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