In this page calculus application problem1 we are going to see solution of the first problem of the topic application problems in calculus.calculus application problem1
A missile fired from ground level rises x meters vertically upwards in t seconds and x = 100t-25t²/2. Find (i) the initial velocity of the missile ,(ii) the time when the height of the missile is a maximum (iii) the maximum height reached (iv) the velocity with which the missile strikes the ground.
Here the variable "x" represents distance taken by the missile and "t" represents times in seconds
The distance covered by the missile in "t" seconds
x = 100t-25t²/2
(i) The initial velocity of the missile
To find the initial velocity,we have to find the value of dx/dt, for that we have to apply 0 for t.
to find dx/dt let us differentiate the given function with respect to "t"
dx/dt = 100 (1) - 25 (2t)/2
= 100 - 25 t
put t = 0,to find the initial velocity
= 100 - 25 (0)
= 100 meter/seconds
(ii) the time when the height of the missile is a maximum
when the missile reaches its maximum height, then velocity of the missile will become zero.
that is dx/dt = 0
100 - 25 t = 0
- 25 t = -100
t = 100/25
t = 4 seconds
(iii) the maximum height reached
from the answer of the previous question we come to know that the missile reaches its maximum height in 4 seconds. From this to find the maximum height we have to apply t = 4 in the given equation. Because we need to find the value of x.
x = 100 (4) - 25 (4)²/2
= 400 - 25 (16)/2
= 400 - 25 (8)
= 400 - 200
= 200 meter
(iv) the velocity with which the missile strikes the ground
when the missile strikes the ground,the distance will be zero.
100 t - 25 t²/2 = 0
(200 t - 25 t²)/2 = 0
200 t - 25 t² = 0
25 t (8 - t) = 0
t = 0 , t = 8
t = 0 is not possible.So let us take t = 8
velocity at t = 8
dx/dt = 100 - 25 t
= 100 - 25 (8)
= 100 - 200
= -100 meter/second calculus application problem1