Question

 A speedboat starts from rest and accelerates at +2.01 m/s²? for 6.90 s. At the end of this time, the boat continues for an additional 5.80 s with an acceleration of +0.518 m/s². Following this, the boat accelerates at-1.49 m/s² for 8.15 s. 

(a) What is the velocity of the boat at t = 20.85 s? 

(b) Find the total displacement of the boat. 

Answer 1

(a) Given in the problem -

+2.01 m/s^2 for 6.90 s

Apply the equation -

v=u+at

=> v1 =0 + 2.01 * 6.90 = 13.87 m/s

s=u*t +(1/2)*a*t^2

s1 = 0 + 0.5 * 2.01 * 6.90^2 = 47.85 m

Again -

0.518 m/s2 for 5.80 s

v2 = 13.87 + 0.518 * 5.80 = 16.87 m/s

s2 = 10.73 * 5.80 + 0.5 * 0.518 * 5.8^2 = 62.23 + 8.71 = 70.94 m

Again -

-1.49 m/s^2 for 8.15 s

So -

v3 = 16.87 - 1.49 * 8.15 = 4.73 m/s

Therefore, velocity of boat at t = 20.85 s is -

v3 = 4.73 m/s (Answer)

And,

s3 = 16.87 * 8.15 - 0.5 * 1.49 * 8.15^2 = 137.49 - 49.48 = 88.01 m

(b) Total displacement of the boat -

s = s1 +s2+s3
= 47.85 m + 70.94 m + 88.01 m = 206.8 m (Answer)