Question

A motorcycle starts from rest with an initial acceleration of 3 m/s^2, and the acceleration then changes with the distance s as shown. Determine the velocity v of the motorcycle when s=200 m. At this point, also determine the value of the derivative dv/ds.

Answer 1

Follows are the solution to this question:

Step-by-step explanation:

Calculating the area under the curve:

A = as

`=(1)/(2)(3 +6 (m)/(s^2))(100 \ m)+ (1)/(2)(6+4 (m)/(s^2))(100 m) \\\\=(1)/(2)(9 (m)/(s^2))(100 \ m)+ (1)/(2)(10(m)/(s^2))(100 m) \\\\=(1)/(2)(900 (m^2)/(s^2))+ (1)/(2)(1,000(m^2)/(s^2)) \\\\=(450 (m^2)/(s^2))+ (500(m^2)/(s^2)) \\\\= 950 \ (m^2)/(s^2)`

Calculating the kinematics equation:

`\to v^2 = v^2_(o) + 2as\\\\`

`=0+ √(2as)\\\\ = √(2(A))\\\\= \sqrt{2(950 (m^2)/(s^2))}\\\\= 43.59 (m)/(s)`

Calculating the value of acceleration:

`\to a= (dv)/(dt)`

`=(dv)/(ds)((ds)/(dt)) \\\\=v(dv)/(ds)\\\\\to (dv)/(ds)=(a)/(v)`

`\to (dv)/(ds) =(4 (m)/(s^2))/(43.59 (m)/(s)) \\\\`

`=(0.092)/(s)`