Mechanics : Equations of motion.

In summary, the motorcyclist's velocity is not affected by the time it takes to travel a certain distance.
  • #1
Maatttt0
37
0

Homework Statement



A motorcyclist starts from rest at a point O and travels in a straight line. His velocity after t seconds is vms^-2, for 0 =< t =< T, where v = 7.2t - 0.45t^2. The motorcyclist's accelaration is zero when t = T.

Find the value of T.

Homework Equations



---

The Attempt at a Solution



S = X
U = 0
V = 7.2t - 0.45t^2
a = 0
t = t

v = u + at
7.2t^2 - 0.45t = 0
0.45t(t - 16) = 0
Therefore t = 0 or 16

But the answer is 8.. I believe I'm getting the accleration part incorrect but I just cannot spot it.
Please help - thank you :)
 
Last edited:
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  • #2
The expression for v you used in step 3 doesn't match the problem statement. You swapped terms.
 
  • #3
Ahh apologies - I mistyped it. The second line is supposed to have the ^2 on the 0.45t :(
 
  • #4
Maatttt0 said:

Homework Statement



A motorcyclist starts from rest at a point O and travels in a straight line. His velocity after t seconds is vms^-2, for 0 =< t =< T, where v = 7.2t - 0.45t^2. The motorcyclist's accelaration is zero when t = T.

Find the value of T.

Homework Equations



---

The Attempt at a Solution



S = X
U = 0
V = 7.2t - 0.45t^2
a = 0
t = t

v = u + at
This is only true for a constant acceleration- and a constant acceleration gives a linear velocity, not quadratic as you have here.

7.2t^2 - 0.45t = 0
0.45t(t - 16) = 0
Therefore t = 0 or 16
All you have done here is solve for V= 0, not acceleration.

But the answer is 8.. I believe I'm getting the accleration part incorrect but I just cannot spot it.
Please help - thank you :)
The acceleration is the derivative of the velocity function. Take the derivative of [itex]7.2t^2- 0.45t[/itex] and set that equal to 0.
 
  • #5
Thank you for the reply. I remember my teacher had mention not to attempt that question as he had not covered it in class yet. I read up about it and you're help has reinforced my understanding.

Thanks again. :)
 

Related to Mechanics : Equations of motion.

1. What are the three equations of motion?

The three equations of motion are:
1. vf = vi + at (equation for calculating final velocity)
2. xf = xi + vit + ½at^2 (equation for calculating final position)
3. vf^2 = vi^2 + 2a(xf - xi) (equation for calculating final velocity without time)

2. What is the difference between displacement and distance?

Displacement is the shortest distance between an object's starting and ending point, while distance is the total amount of ground covered by the object. Displacement is a vector quantity, meaning it has both magnitude and direction, while distance is a scalar quantity, meaning it only has magnitude.

3. How do you calculate acceleration?

Acceleration can be calculated using the equation a = (vf - vi)/t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time.

4. Can the equations of motion be used for any type of motion?

No, the equations of motion are specifically designed for motion with constant acceleration. If the acceleration is not constant, other equations and principles, such as calculus and the laws of motion, must be used.

5. What is the significance of the letter "g" in the equations of motion?

The letter "g" represents the acceleration due to gravity, which is approximately 9.8 m/s^2 on Earth. This value is used in the equations of motion to calculate the effect of gravity on an object's motion.

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