Question

Please Help!!!! ASAP!!!

A remote-controlled car is moving in a vacant parking lot. The velocity of the car as a function of time is given by υ⃗ =[5.00m/s−(0.0180m/s3)t2]i^ + [2.00m/s+(0.550m/s2)t]j^.

a. What is the magnitude of the acceleration of the car at t = 6.23 s ?

I already know that i need to use the derivative of the velocity equation given but I don't know what to do from there.

Answer 1

`0.594 m/s^2`

Step-by-step explanation:

The acceleration is the derivative of the velocity. Therefore, we need to differentiate each component of the velocity to find the components of the acceleration first.

The velocity along the x-axis is:

`v_x = 5.00 - 0.0170 t^2`

Differentiating,

`a_x = -2(0.0180)t = -0.0360 t`

Similarly, the velocity along the y-axis is:

`v_y = 2.00 + 0.550 t`

Differentiating,

`a_y = +0.550`

So now we have the components of the acceleration; substituting t = 6.23 s, we get:

`a_x = -0.0360 (6.23)=0.224 m/s^2\\a_y = 0.550 m/s^2`

And so the magnitude of the acceleration is

`a=√(a_x^2+a_y^2)=√((0.224)^2+(0.550)^2)=0.594 m/s^2`