ce amath -- integration

1. A stone falling freely from a height 20m above the ground has ca constant accleration of 10 m/s^2. What is its velocity just before it reaches the ground.

the acceleration a= 10 so the velocity v= ∫a dt =10t +C
when t=0 , v=0
so 0= 10(0) +C
C=0

displacement of the stone d = ∫v dt
=∫(10t )dt
=5t^2 +D
when t=0 , d=0
so 0= 5(0)^2 +D
D=0

so the relation between d and t is
d =5t^2

when the stone is just touching the ground displacement =20
20 =5t^2
t^2 =4
t =2 or -2(rej.)

velocity of the stone at that time
=10(2)
=20m/s

2.A man drives from town A to town B on a staright road without stopping . His velocity at time t hours after leaving A is (60t-3t^2)km/h. Find

(a)his travelling time.
since he drive without stopping
when he arrive at B the velocity should be maximum
so
d(60t-3t^2)/dt =60-6t
(60-t)
d^2(60t-3t^2)/dt^2 = -6

put d(60t-3t^2)/dt =0
we get the maximum velocity
60-6t=0
t=10 hours

his travelling time=10 hours

(b) the distance between town A and town B.

the displacement D from town A to town B is given by ∫(60t-3t^2)dt
=30t^2 -1.5t^3 +K
when t=0, D=0
0=30(0)^2 -1.5(0)^3 +K
K=0
so D=30t^2 -1.5t^3
when he arrive at B, t=10
D=30(10)^2 -1.5(10)^3
=1500km
the distance between town A and town B is 1500km


(c) the maximum velocity of the car
put t =10
(60t-3t^2)= (60*10-3(10)^2)
=300km/h