Calculating Time for Constant Acceleration: Motorcycle Velocity Change

In summary, the conversation discusses a problem involving a motorcycle with a constant acceleration of 2.80 m/s2. The question asks for the time required for the motorcycle to change its speed from 40.9 to 50.9 m/s and from 70.9 to 80.9 m/s. The correct method to solve this problem is by using the kinematic formula Vf = Vi + a*t and solving for t by isolating the variable.
  • #1
Erenjaeger
141
6

Homework Statement


A motorcycle has a constant acceleration of 2.80 m/s2. Both the velocity and acceleration of the motorcycle point in the same direction. How much time is required for the motorcycle to change its speed from (a)40.9 to 50.9 m/s, and (b)70.9 to 80.9 m/s?
[/B]

Homework Equations


Vf = Vi + a*t[/B]

The Attempt at a Solution


using that kinematic formula,
for a) 50.9=40.9+2.8*t
then divide both sides by 40.9+2.8 to solve for t
answer = 4.04[/B]
and same thing with b) but in both cases the answer is coming up as wrong. what have i done wrong?
 
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  • #2
Erenjaeger said:

Homework Statement


A motorcycle has a constant acceleration of 2.80 m/s2. Both the velocity and acceleration of the motorcycle point in the same direction. How much time is required for the motorcycle to change its speed from (a)40.9 to 50.9 m/s, and (b)70.9 to 80.9 m/s?
[/B]

Homework Equations


Vf = Vi + a*t[/B]

The Attempt at a Solution


using that kinematic formula,
for a) 50.9=40.9+2.8*t
then divide both sides by 40.9+2.8 to solve for t
answer = 4.04[/B]
and same thing with b) but in both cases the answer is coming up as wrong. what have i done wrong?
You should check your algebra.

[itex] 50.9 = 40.9 + 2.8t [/itex]

is not the same thing as

[itex] 50.9 = (40.9 + 2.8)t. [/itex]

And for that matter, neither one gives an answer of 4.04 sec. So I'm not really sure what happened there.

(Hint, you can't just add 40.9 plus 2.8 that way. That's not the way algebra works.)
 
  • #3
collinsmark said:
You should check your algebra.

[itex] 50.9 = 40.9 + 2.8t [/itex]

is not the same thing as

[itex] 50.9 = (40.9 + 2.8)t. [/itex]

And for that matter, neither one gives an answer of 4.04 sec. So I'm not really sure what happened there.

(Hint, you can't just add 40.9 plus 2.8 that way. That's not the way algebra works.)
oh right, so basically i have to get 2.8t on its own and then divide by 2.8??
 
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  • #4
Erenjaeger said:
oh right, so basically i have to get 2.8t on its own and then divide by 2.8??
Yes, that is the correct way. :smile:
 
  • #5
collinsmark said:
Yes, that is the correct way. :smile:
cool thanks!
 
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1. What is constant acceleration?

Constant acceleration is a type of motion where an object's velocity changes by the same amount over equal time intervals. This means that the object's speed increases or decreases by the same amount over a certain period of time.

2. How is constant acceleration different from uniform motion?

Uniform motion is when an object's velocity remains constant, meaning it does not change over time. In contrast, constant acceleration involves a change in velocity, either speeding up or slowing down, over a period of time.

3. What is the equation for calculating constant acceleration?

The equation for calculating constant acceleration is a = (vf - vi) / t, where a is the acceleration, vf is the final velocity, vi is the initial velocity, and t is the time interval.

4. How does constant acceleration affect an object's motion?

Constant acceleration affects an object's motion by changing its velocity over time. This means that the object's speed and direction will change, causing it to move in a curved path.

5. What are some real-life examples of constant acceleration?

Some real-life examples of constant acceleration include objects falling due to gravity, a car speeding up or slowing down, and a roller coaster moving along its track. In all of these cases, the object's velocity is changing by the same amount over equal time intervals, resulting in constant acceleration.

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