Question 1 A certain automobile manufacturer claims that its...

Question 1 A certain automobile manufacturer claims that its deluxe sports car will accelerate from rest to a speed of 180km/hr in 10 s. a. Define your variables (1 mark) b. Determine the average acceleration of the car in m/s² (1 mark) c. If the car moves at a constant acceleration, find the distance (m) that the car travels in the first 10 s. (1 mark) d. What is the speed of the car 12.0 s after it begins its motion if it can continue to move at the same acceleration? Give this speed in km/hr. Use the equations: Average acceleration, $a_{avg}$ is given by $\frac{\Delta V}{\Delta t} = \frac{V_f - V_i}{t_f - t_i}$ Where initial velocity of the car, $V_i$, final velocity of the car, $V_f$ and time elapsed, $\Delta t$. At constant acceleration, $a_{avg}$, the distance, $\Delta x$, that is covered by the car is given by the equation below. $\Delta x = \frac{1}{2}*(V_i + V_f)*t$

Transcript

00:01 Hello, in the question we have given that a car is moving in a linear path with accelerates from rest at constant acceleration for a distance of 300 meters.

00:12 So for first 300 meters, the acceleration is constant and it is starting from rest.

00:18 So it then maintains velocity for 15 seconds.

00:24 So for 15 seconds, the velocity is constant.

00:27 Then the driver applies brakes.

00:31 So and the this velocity is given.

00:36 So velocity equation is given that is v is equal to 30 cos t.

00:40 So we say that where t is equal to 0 seconds when the brakes are applied.

00:47 So now in the first part, we have to find the time taken for the car to come to stop.

00:52 So now when the brakes is applied, the velocity is given as v is equal to 30 cos t.

00:59 So now the finally it is coming to stop.

01:02 So when the car is stops, this velocity should be equal to 0.

01:07 So we will put this velocity equal to 0, which implies that cos of t should be equal to 0.

01:14 So now when this cos can be 0, when this t, the angle should be equal to pi by 2.

01:20 So cos 90 is 0.

01:21 So t should be equal to pi by 2.

01:23 So from here, we get this time as equal to 1 .57 seconds.

01:27 So this will be the answer for the first part.

01:32 Now moving to the second part, what they are saying, find the acceleration of the car during first 300 meters.

01:39 So we have to find out acceleration of car during the first 300 meters.

01:44 Now see the velocity before applying the brake is v is equal to 30 cos t.

01:50 So now before applying the brakes, so when the brakes are applied, so this velocity will be equal to, so at t is equal to 0.

02:00 So we will just put t over here 0.

02:03 So cos of 0 is equal to 1.

02:06 So this velocity becomes 30 meters per second.

02:08 So before this brakes is applied, this is the velocity.

02:12 Now this velocity is constant for 15.

02:14 So for this 15 seconds, so we can say that initial velocity is 0 because it starts from rest.

02:22 Then the final velocity is 30 at when it starts moving with a constant velocity.

02:30 So it is 30 meters per second.

02:33 Then the distance is 300 meters.

02:36 So we can use this equation vf square is equal to vi square plus 2s.

02:41 So from here, this initial is 0.

02:44 So this goes away.

02:45 So we get acceleration as equal to 30 square divided by 2 into 300.

02:49 And after calculating, we get the acceleration as 1 .5 meters per second square.

02:55 So now in the third part, what we have to do is we have to find the total distance traveled from rest to stop.

03:02 So we have to calculate the total distance...

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