A cyclist starts from rest from her school and accelerates uniformly to reach Bus Stop A at a speed of 6 m/s. She then cycles at constant speed for some time till she reaches Bus Stop B. Then, she decelerates uniformly and comes to rest when she reaches her house. The distance between Bus Stop A and B is half of the total distance covered. If the acceleration is twice the deceleration and her total travel time is 30 seconds, find the acceleration.
Answers
Answer:
Step-by-step explanation:
deceleration from B to home = a
acceleration from school to A = 2a
let time (school to A) =t = dv/2a
we are going to write everything in terms of t to solve this,
t = modulus of delta velocity /acceleration = dv/2a
time from B to home = dv/a= 2t
Distance ( school to A) = Vavg *t = (0+6)*t/2= 3t
Distance ( B to home)= Vavg *2t= 2*Vavg*t= 2*3t=6t
Distance (AB) = half of Total distance = [ Distance (school to A) + Distance (B to home)] = 3t + 6t = 9t
Distance (AB) = 9t , so time (AB) = Distance / velocity (6m/s) = 9t/6
Total time = sum of all time intervals = 30s
hence t + 9t/6 + 2t = 30
when we solve the above we get t= 20/3s
hence acceleration (school to A) = dv/t
= (6-0)/(20/3) = 6*3/20 = 18/20 =9/10 = 0.9 m/s^2
deceleration =a = 0.9/2=0.45m/s^2
ans. 0.9m/s^2