Question

6.
The engine of a motorcycle can produce a maximum acceleration
m/s. Its brakes can produce a maximum retardation 10 m/
What is the minimum time in which it can cover a distance of 1.5km
sec
​

Answer 1

Let’s say that the bike starts from zero and accelerates at 5m/s^2 for t secs until it reaches v m/s

Then (v-0)/t = 5 or v=5t , and -v = -5t

If T is the total time then there is T-t secs left to stop using a deceleration of

-10 m/s^2 ,so

(0-v)/(T-t) =-10

-v = -10T +10t but -v = -5t , so

-5t = -10T + 10t

10T = 15t

T = 3t/2

Now use the piece of information given the distance 1500m

[1/2] v T = 1500

v T =3000 . Note we have v and T in term of one [1] unknown t

[5t][3t/2]=3000

15t^2 = 6000

t^2 = 400

t=20

So T = 3t/2= 3[20]/2 = 30 secs is the shortest time to complete the jorumey.

Answer 2

answer is 30..

see the attached image...