(13%) Problem 1: Referring to the figure, a car travels a...

(13%) Problem 1: Referring to the figure, a car travels a distance d = 21.1 m in the positive x direction in a time t1 = 22.7 s, at which point the car brakes, coming to rest in t2 = 5.08 s. 25% Part (a) What was the car's average velocity in the horizontal direction, in meters per second, during the t1 time period? 25% Part (b) Assuming the car started from rest and moved with a constant acceleration, what was the acceleration, in meters per squared second, during time interval t1? 25% Part (c) What was the car's instantaneous velocity in the horizontal direction, in meters per second, when it began braking? 25% Part (d) Using the result from Part (c), what was the car's horizontal component of acceleration, in meters per squared second, during the braking period?

Transcript

00:01 Hello, in the question we have given that a car travels a distance of d is equal to 21 .1 meters.

00:06 So, let us write what are the known quantities.

00:09 So, d is given that is 21 .1 meters.

00:12 So, in the positive, so this car travels this distance d in positive direction in time t1.

00:18 So, the t1 is given that is 22 .7 seconds at which point car brakes coming to rest in t2.

00:28 So, t2 is given that is 5 .08 seconds.

00:32 So, now see at this distance 21 .1 meters, the brakes are applied and it takes this time 5 .08 seconds to come to rest for this car.

00:45 So, now there are four parts.

00:48 So, in the first part, they are asking us to find out the average velocity in time t1.

00:52 So, in the time period t1.

00:54 So, v average will be equal to d by t1 because this is the distance traveled in time t1.

01:00 So, it will be 21 .1 divided by t1 is 22 .7.

01:06 So, if we calculate from here, so v average will we will get it as 0 .9295 meters per second, which we can write it approximately as equal to 0 .93 meters per second.

01:22 So, this is the average velocity which we get from here.

01:25 Now, moving to the next part.

01:27 So, they are asking us to find out the acceleration.

01:31 So, assuming the car is car started from rest that is u is equal to 0.

01:36 So, and moved with a constant acceleration in meters per second.

01:41 So, what we have to do is we during the time interval t1.

01:44 So, we have to find out this acceleration only.

01:47 So, we have to find out acceleration.

01:49 So, we will use this equation s is equal to u t1 plus half a t1 square because in this time period only we are we have been asked to find out the acceleration.

02:00 So, this term will go to 0 because u is 0, the initial velocity is 0.

02:05 So, from here a will be equal to 2s divided by t1 square.

02:10 So, let us plug the value...

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