Calculating Time for Constant Acceleration: Motorcycle Velocity Change

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Erenjaeger
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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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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.)
 
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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collinsmark said:
Yes, that is the correct way. :smile:
cool thanks!
 
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