The engine of a motorcycle can produce a mai mdacceleration 5 m/s. Its

1.	The engine of a motorcycle can produce a mai mdacceleration 5 m/s. Its brakes can produce a manretardation 10 m/s. What is the minimum time in wcan cover a distance of 1.5 km[Pb. PM la(a) 30 sec(b) 15 sec(c) 10 sec(d) 5 sec
Answer» 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 journey given the conditions. therefore answer is 30secs

The engine of a motorcycle can produce a mai mdacceleration 5 m/s. Its brakes can produce a manretardation 10 m/s. What is the minimum time in wcan cover a distance of 1.5 km[Pb. PM la(a) 30 sec(b) 15 sec(c) 10 sec(d) 5 sec

Answer»

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 journey given the conditions.

therefore answer is 30secs

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