The given function represents the position of a particle traveling along a horizontal line. s(t) = 2t3 − 3t2 − 36t + 6 for t ≥ 0
1 answer:
1) The velocity of the particle is given by the derivative of the position. So, if we derive s(t), we get the velocity of the particle as a function of the time:
2) The acceleration of the particle is given by the derivative of the velocity. So, if we derive v(t), we get the acceleration of the particle as a function of the time:
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1) The law of motion of the projectile is
To find the velocity, we should compute the derivative of h(t):
So now we can calculate the speed at t=2 s and t=4 s:
The negative sign in the second speed means the projectile has already reached its maximum height and it is now going downward.
2) The projectile reaches its maximum height when the speed is equal to zero:
So we have
And solving this we find
3) To find the maximum height, we take h(t) and we just replace t with the time at which the projectile reaches the maximum height, i.e. t=2.30 s:
4) The time at which the projectile hits the ground is the time at which the height is zero: h(t)=0. So, this translates into
This is a second-order equation, and if we solve it we get two solutions: the first solution is negative, so we can ignore it since it's physically meaningless; the second solution is
And this is the time at which the projectile hits the ground.
5) The velocity of the projectile when it hits the ground is the velocity at time t=4.68 s:
with negative sign, because it is directed downward.
To find the velocity, we should compute the derivative of h(t):
So now we can calculate the speed at t=2 s and t=4 s:
The negative sign in the second speed means the projectile has already reached its maximum height and it is now going downward.
2) The projectile reaches its maximum height when the speed is equal to zero:
So we have
And solving this we find
3) To find the maximum height, we take h(t) and we just replace t with the time at which the projectile reaches the maximum height, i.e. t=2.30 s:
4) The time at which the projectile hits the ground is the time at which the height is zero: h(t)=0. So, this translates into
This is a second-order equation, and if we solve it we get two solutions: the first solution is negative, so we can ignore it since it's physically meaningless; the second solution is
And this is the time at which the projectile hits the ground.
5) The velocity of the projectile when it hits the ground is the velocity at time t=4.68 s:
with negative sign, because it is directed downward.
To calculate the initial velocity of the bike, we use the following equation
.
or
Here, u is initial velocity, v is final velocity, t is the time and d is the distance covered by bike.
Given, ,
and
.
Substituting these values in above equation, we get
.
Thus, the initial velocity of the bike is 1.2 m/s.