Question 9 A car moves along a straight line whose equation...

Question 9 A car moves along a straight line whose equation of the motion is given by: $s = 12t + 3t^2 - 2t^3$ where 's' is in meters and 't' is in seconds. Calculate the velocity and acceleration at start as well as acceleration when the velocity is zero. [CLO2/PLO2,C4] (10 marks) The equation of motion of a particle moving in a straight line is given by: $s = 18t^2 + 3t^3 - 2t^4$ where 's' is in meters and 't' is in seconds. Find the velocity and acceleration at start, time when particle reaches its maximum velocity, and maximum velocity of the particle. [CLO2/PLO2,C4] (10 marks) END OF QUESTION PAPER

Transcript

00:01 Hello students, in this question, in the first part of the question given as the displacement in function of time is 12t plus 3t square minus 2t cube.

00:19 Now, we have to calculate the velocity.

00:22 So, velocity is equals to ds by dt is equal to the 12 plus 6t minus 6t square, equation number one.

00:34 In this question, we have to calculate the velocity at time initially initial time.

00:40 So, at time t is equals to zero, the velocity will be equal to the 12 meter per second.

00:48 Next, we have to calculate the acceleration.

00:51 So, acceleration is equals to dv by dt or we can write d2 s by dt square.

01:01 So, this will be equal to the 6 minus 12t.

01:07 But the acceleration, we have to calculate the acceleration when velocity was equal to the zero.

01:13 When velocity was equal to the zero, then from the equation number one, then from equation one, when velocity is zero, so this will be equal to the 12 plus 6t minus 6t square.

01:30 This is a quadratic equation.

01:33 So, after solving this equation, we will get the time is equals to two and minus one second.

01:40 So, minus is not exist negative time.

01:44 So, we will take the time t is equals to two second and we will put the value of this two second in the equation number.

01:57 And after calculation, we will get the acceleration is equals to 6 minus 12 into 2.

02:06 So, this will be equal to the minus 18 meter per second square.

02:12 In the next part of the question, given as the displacement is equal to 18t square plus 3t cube minus 2t to the power 4.

02:33 Now, again, we will calculate the velocity is equals to ds by dt...

Question

Please give Ace some feedback