Question

A cab driver picks up a customer and delivers her 3.50 km away, on a straight route. The driver accelerates to the speed limit and, on reaching it, begins to decelerate all at once. The magnitude of the deceleration is two times the magnitude of the acceleration. Find the lengths of the acceleration and deceleration phases.

Answer 1

Given

the distance is s = 3.50 km = 3500 m

in the accelating phase

the distance travelled by the cab when it reaching the speed limit from rest is  

using equations of motions

V^2-u^2 = 2a*s1

v^2 -0 = 2*a*s1

s1 = v^2/2*a

decelerating phase  

starting from maximum velocity and final velocity is zero , and deceleration if -2a

V^2-u^2 = 2a*s2

-v^2 = -2*2a*s2

s2 = v^2/4*a

here comparing the distances s1,s2 , s1 = 2*s2

and the total distance is s = s1+s2 = 3.5km

3.5 km = 2s2+s2 = 3s2

s2 = 3.5/3 km = 1.16667 km

and s1 = 2*s2 = 2*1.16667 km = 2.33334 km

the lengths of acceleration phase is s1 = 2.33334 km ,

the lengths of deceleration phase is s2 = 1.16667 km ,