Calculating Distance Traveled: Motion Exam Question Solution in 7.4 Seconds

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SUMMARY

The discussion focuses on calculating the distance traveled by a cyclist accelerating at 1.1 m/s² from an initial speed of 4.2 m/s over a duration of 7.4 seconds. The correct formula to use is s(t) = (1/2)at² + v₀t + s₀, where 's' represents distance, 'a' is acceleration, 'v₀' is the initial velocity, and 't' is time. The final calculated distance is confirmed to be 61 meters, highlighting a common mistake of misapplying the formula by incorrectly squaring the entire acceleration term.

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1. A Cyclist traveling at a speed of 4.2 ms^-1 accelerates at 1.1 ms^-2 in a time of 7.4 seconds. What is the distance travelled?


2. Homework Equations ?



3. The Attempt at a Solution : The answer is 61m but i don't know how to get that answer
 
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I would expect you to have seen s(t)=\frac{1}{2}at^2+v_0t+s_0?
 
Yeh thanks i got it
 
When i was doing it i was squaring the whole of the at instead of just the t
 
I see. You should really put that in the relevant equation section, that way people reading this forum could have seen what went wrong right away.
 
Yh sorry this was my first post but thanks anyway
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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