Vectors - Displacement to Accelleration

In summary, the equations given in the question were x=5t^3-6t^2 and y=3t^-1+4t^4, with a requested magnitude of acceleration at t=1s. After finding the second derivatives of both equations, the incorrect simplification of y led to an incorrect resultant acceleration of 19.69 m/s^2 instead of the correct answer of 57 m/s^2.
  • #1
cryodynamics
4
0

Homework Statement


An object moves with its position obeying x=5t^2-6t^2 and y=3t^-1+4t^4. What is the magnitude of its acceleration at t=1s;
a)57 m/s^2
b)34 m/s^2
c)26 m/s^2
d)44 m/s^2
e)0 m/s^2

Homework Equations


x=5t^3-6t^2
y=3t^-1+4t^4

The Attempt at a Solution


Second derivative of the displacement x is;
ax=30t-12 m/s^2
Second derivative of the displacement y is;
ay=8t m/s^2

t=1 -> ax
ax = 18

t=1 -> ay
ay = 8

Resultant acceleration = sqrt(18^2+8^2) = 19.69

Which is wrong according to the answer I found online which was a) 57 m/s^2. Anyone show me where I've messed up or is the answer I found wrong?
 
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  • #2
cryodynamics said:

Homework Statement


An object moves with its position obeying x=5t^2-6t^2 and y=3t^-1+4t^4. What is the magnitude of its acceleration at t=1s;
a)57 m/s^2
b)34 m/s^2
c)26 m/s^2
d)44 m/s^2
e)0 m/s^2


Homework Equations


x=5t^2-6t^2
y=3t^-1+4t^4

The Attempt at a Solution


Second derivative of the displacement x is;
ax=30t-12 m/s^2
Second derivative of the displacement y is;
ay=8t m/s^2

t=1 -> ax
ax = 18

t=1 -> ay
ay = 8

Resultant acceleration = sqrt(18^2+8^2) = 19.69

Which is wrong according to the answer I found online which was a) 57 m/s^2. Anyone show me where I've messed up or is the answer I found wrong?

The x expressions don't match up, but I guess you actually meant x=5t^3-6t^2

I can't imagine what you thought the y expression was ??
 
  • #3
PeterO said:
The x expressions don't match up, but I guess you actually meant x=5t^3-6t^2

I can't imagine what you thought the y expression was ??

Ahh yes sorry about that, I'll fix that up. What do you mean by 'I can't imagine what you thought the y expression was?' That equation for y is the exact one given?
 
  • #4
Can you check that

x=5t^2-6t^2 and y=3t^-1+4t^4

are correct.
 
  • #5
x=5t3-6t2
y=3t-1+4t4
Are the the correct equations given in the question.
 
  • #6
cryodynamics said:
x=5t3-6t2
y=3t-1+4t4
Are the the correct equations given in the question.

OK .. if that is the expression for y, then 8t is not the second derivative.
 
  • #7
PeterO said:
OK .. if that is the expression for y, then 8t is not the second derivative.

Ahh I got the answer this time, my problem was I thought I could simplify y=3t-1+4t4 to 4t3/3 I now realize what a rookie error that was lol. Thanks for your help everyone.
 

Related to Vectors - Displacement to Accelleration

1. What is a vector?

A vector is a mathematical quantity that has both magnitude and direction. It is commonly represented graphically as an arrow, with the length of the arrow indicating the magnitude and the direction of the arrow indicating the direction of the vector.

2. What is displacement?

Displacement is the change in position of an object from its original position to its final position. It is a vector quantity, meaning it has both magnitude (distance) and direction. It is typically measured in units such as meters or kilometers.

3. How is displacement related to acceleration?

Displacement and acceleration are related through the equations of motion. Displacement is the integral of acceleration, meaning that acceleration is the rate of change of displacement over time. This relationship is described by the equation a = Δv/Δt, where a is acceleration, Δv is the change in velocity, and Δt is the change in time.

4. What is the difference between average and instantaneous acceleration?

Average acceleration is the average rate of change of velocity over a given time interval. It is calculated by dividing the change in velocity by the change in time. Instantaneous acceleration, on the other hand, is the acceleration at a specific moment in time. It is calculated by taking the limit as the time interval approaches zero.

5. How can vectors be used to calculate acceleration?

Vectors are used in the study of displacement to acceleration by representing both the initial and final positions of an object, as well as the direction and magnitude of its movement. By analyzing these vectors, scientists can calculate the acceleration of an object using the equations of motion and determine its direction and magnitude.

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