A car starts at point A from rest and accelerates at a=3*s...

A car starts at point A from rest and accelerates at a=3*s where a is in m/s$^2$ and s is in m. It then maintains a constant speed from B to C. Find: • The speed of the car at B • The time it takes for the car to get from B to C (BC is a 45 degree arc) • The magnitude of the acceleration at C (on the curve shown) $r = 10 m$ $27 m$ A

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00:04 Car is moving from point a to point b and the length a b is given as 885 meters and it is also given that the car starts from a and then accelerate at the constant of a meter per second square and it's also given that the speed of car at a oh no at when it reaches the point after acceleration when it reaches some point it is constant and that constant speed is at after some point is we let's suppose this is v is equal to 15 meter per second so it start from a accelerate till point some point and then reaches the constant speed and that speed is 15 meter per second and the acceleration is a meter per second square and also it says that it accelerates for time t by three seconds so this is a time for which is accelerating now then it's maintaining the maintaining the speed of 15 meter per second for total time of t seconds and then it says that it's de -accelade at the constant rate of 2 .5 meter per second square and that's second acceleration let's call that as a two and then stopping at b so here we have to first find what is the time for which the car decelerates and then we have to sketch the speed time graph of the complete motion of the car and then we have to find the value of t and the value value of a and then we have to grab the acceleration time graph so we'll start with finding the the first part which is the time for which the car is de -acceleration de -accelating so that is equal to t over three so here the de -acceleration rate is given as 2 .5 meters per second square so week and before that it is moving at the speed of 15 meter per second which is a constant so we'll be having v1 as 15 meter per second and since the car is stopping so v2 would be equal to zero because zero meter per second so using the formula v2 is equal to v1 plus a let's call t3 because t by 3 is taken as t3 so here we'll be getting 0 is equal to 15 since this is a d acceleration so is taken as minus 2 .5 is the acceleration or d a is a deacceleration rate so 2 .5 t3 and then we sort that for so t3 would be equal to six seconds so that would be the answer for a so total for six second the car is de -accelerating then we'll find what is the distance that is stable during this time of de -excellation let's call that distance as s 3 so here we'll be using this formula v square minus u squared is equal to 2 as so final velocity is 0 minus 15 is the initial velocity 15 square minus 2 into 2 .5 in times s 3 so we'll solve that for s 3 and that is equal to 45 meters so we can say that let's call this distance as so the total distance would be let let's call this distance is 1 plus s2 and that is equal to 885 minus 45 which is equal to 840 meters here s1 is the distance when the distance travel when the car is accelerating and s2 is the distance the constant distance or when it's traveling with the constant speed so it is a distance…

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