Question

A rock thrown vertically upward from the surface of the moon at a velocity of 32​m/sec reaches a height of sequals32 t minus 0.8 t squaredmeters in t sec.a. Find the​ rock's velocity and acceleration at time t.b. How long does it take the rock to reach its highest​ point?c. How high does the rock​ go?d. How long does it take the rock to reach half its maximum​ height?e. How long is the rock​ aloft?

Answer 1

Answer:

A) v(t) = (32 - 1.6t) m/s

a(t) = -1.6 m/s²

B) t = 20 seconds

C) 320 m

D) t = 5.85 seconds going up and 34.14 seconds going down

E) 40 seconds

Explanation:

Height is given by the equation;

S(t) = 32t - 0.8t²

A) Velocity after time t is gotten from first derivative of the distance.

Thus;

v(t) = dS/dt = 32 - 1.6t

Acceleration at time t is gotten from derivative of the velocity.

Thus;

a(t) = d²S/dt² = -1.6 m/s²

B) At highest point, velocity is zero.

Thus;

32 - 1.6t = 0

1.6t = 32

t = 32/1.6

t = 20 seconds

C) To find how high the rock goes, it means we are looking for maximum height.

This will be at t = 20 seconds.

Thus;

S(20) = 32(20) - 0.8(20)²

S(20) = 640 - 320

S(20) = 320 m

D) we want to find the time it will take to reach half its maximum height.

since maximum height is 320 m, then half the maximum height is; S_½ = 320/2 = 160

Thus;

160 = 32t - 0.8t²

0.8t² - 32t + 160 = 0

Using quadratic formula, we will get;

t = 5.85 seconds going up and 34.14 seconds going down

E) time the rock is aloft = twice the time it took to reach maximum height.

Thus; t_aloft = 2 × 20 = 40 seconds