A car starts from rest & again comes to rest after travelling 200 m in a straight line. If its acceleration
and deacceleration are limited to 10 m/s2 & 20 m/s2 respectively then minimum time the car will take to
travel the distance is -
Answers
Answer:
consider that the car accelerates for a time t and decelerates for a time T.
Final velocity of car after time t
v = u + a1t = 0 + 10t
v = 10 t ----- (1)
Now the car will decelerate for time T, and
initial velocity for deceleration = u2 = final velocity after acceleration = v =10t
from the first equation of motion
v after deceleration = u2 -a2T
10t = 20T
t = 2T ---- (2)
Applying the second equation of motion for the time T
s2 = u2T −12a2T^2 = 10 t ×T − 12×20×T2u2T -12a2T2 = 10 t ×T - 12×20×T^2
substituting the value of 't' from equation (2) in the above equation, we have,
s2 = 20T^2 − 10T^2 = 10T^2
s2 = 10T^2
total distance travelled =s1 + s2
20 T^2 + 10T^2 = 200
T = 2.58 s
t = 2T = 2.58 × 2
t = 5.16s
Therefore,
Total time taken to cover the distance of 200m = 2.58+5.16 = 7.74 s