Find Average Acceleration of a Turning Car in 2.0s | Step-by-Step Guide

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To calculate the average acceleration of a car turning from north to east over 2.0 seconds, the initial velocity is represented as 40j (north) and the final velocity as 30i (east). The average acceleration formula a = (v - u) / t is applied, where v is the final velocity and u is the initial velocity. The change in velocity is calculated as (30i - 40j) and converted from km/h to m/s for accuracy. The magnitude of the acceleration vector is determined using the formula sqrt(a^2 + b^2). Detailed steps and unit vector notation are emphasized for clarity in understanding the direction of acceleration.
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Homework Statement


A car traveling north at a speed of 40km/h turns right to head due east at a speed of 30km/h. This change in direction and speed takes 2.0 s. Calculate the average acceleration of the car.


Homework Equations


a = v-u/t


The Attempt at a Solution



Was wondering if you guys could help me with this question. I can find the magnitude quite easily however i cannot find the new direction. Would someone be able to post a detailed guide as to how to get the direction?
 
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Write v and u using unit vectors i and j.
 
Sorry mate I'm kind of new to physics. Would you be able to give me some detailed working out (being annotated would help alot) as to how you come across the acceleration. I no I am asking a lot but its annoying me so much cause i want to learn this before i progress and forget about it
 
Well, uhh...matey, all right.

Choose y-axis toward N and x-axis toward E.

u = 40j and v = 30i.

Average a = change in velocity/time = (v-u)/t = (30i - 40j)/2* 1000/3600 m/s (converting km/h to m/s). Simplify a bit.

You had said that you can find the magnitude quite easily. Still, the magnitude of ai+bj is sqrt(a^2+b^2).
 
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