Driving down the road at a speed of 246 ms you suddenly...

Driving down the road at a speed of 24.6 m/s, you suddenly notice a fallen tree blocking the road a distance of 91.0 m ahead of you. You step on the brake pedal and decelerate at a constant rate. What must the magnitude of your acceleration be so that you will come to a stop 7.3 m in front of the tree? 7.23 m/s^2 3.62 m/s^2 3.33 m/s^2 3.08 m/s^2 An unidentified flying object (UFO) is observed to travel a total distance of 42000 m, starting and ending at rest, over a duration of 2.91 s. Assuming the UFO accelerated at a constant rate to the midpoint of its journey and then decelerated at a constant rate the rest of the way, what was the magnitude of its acceleration? Express your answer in g s , where 1 g = 9.81 m/s^2. 1,011 g s 19,839 g s 9,920 g s 2,022 g s

Transcript

00:01 In the given question we have that on driving down the road at a speed of 24 .6 meters per second, we suddenly notice a fallen tree blocking the road at a distance of 91 .0 meters ahead.

00:14 So we step on the brake and decelerate at a constant rate.

00:17 So we have to calculate the magnitude of the acceleration so that we come at us to stop at a distance of 7 .3 meters.

00:26 So we know that the total distance that is to be covered is.

00:31 Will be equal to x1 minus x2 while decelerating so this will be equal to 91 .0 meters minus 7 .3 meters therefore the distance s that is covered is 83 .7 meters now we need to calculate our acceleration so we will use the canonical equation v square is equal to u square plus 2 a s here v is the final velocity but the final velocity will be equal to 0 when we have to stop the initial velocity we have u square and this is 2a s so the total acceleration a will be equal to minus of u square upon 2s so substituting the values we will have u as 24 .6 meters per second this will be our whole square divided by this is two times the distance covered is 83 .7 meters so a deceleration will be minus 3 .62 meters per second and the magnitude of the deceleration will be 3 .62 meters per second square which is in our second option so this is the first question now let us solve the second one so in this question we have a ufo unidentified flying object is observed to travel a total distance of 42 ,000 meters starting and ending at rest over a duration of time 2 .913 seconds this is the total time we have to assume that ufo is accelerated at a constant rate to the midpoint of its journey and then decelerated at the constant rate and then we have to calculate the magnitude of the acceleration and express it in terms of g which is acceleration due to gravity so first we will find out the time taken when it is accelerating we will use the equation x is equal to u t plus half a t square your x is the total distance as it covered in half of the part so in the accelerated motion we will have the total distance covered as s upon 2 initial acceleration will be initial velocity of it will be equal to 0 and this will be half of a t square so substituting the value and rearranging this we will get t is equal to s upon a root so this is equal to 4 ,000 40 ,000 meters upon acceleration a now we will calculate the time taken for the decelerated part so for the decelerated motion we will use the kinematical equation v is equal to u plus 18 but for decelerated motion we will have acceleration as minus and our final velocity will be equal to zero after it has come to a stop so this will be equal to 0 equal to u minus a t here let this t be equal to t dash to represent that the time taken for the decelerated motion therefore our u is equal to a t dash now we will use again the equation v square is equal to u square plus 2 a s for the decelerated part so again our final velocity will be equal to 0 our initial velocity will be u square and our acceleration will be minus a so this will be minus of 2 times a times s will be s upon 2 so we will have minus of u square equal to as but u that we have already seen is at so this will be minus of a dash t dash equal to 4 2 triple 0 times a which is our acceleration so one of its the acceleration will go and we will get t -dash is equal to root of 4 -2 -300 divided by a but we see that this is the same as the accelerated time that is t so the total time t -dash will be equal to t which is equal to total time which was 2 .91 seconds divided by 2 so both of the times will be equal to 1 .45 seconds therefore our acceleration a which is equal to distance covered divided by the time square is equal to 42 ,000 divided by 1 .455 seconds the whole square...

Question

Please give Ace some feedback