accelerating objects

Question

An object starts at time t = 0 seconds with a velocity of v₀ = +48 m/s and undergoes a constant acceleration of a = -16 m/s².

1) At what time does the speed of the object reach zero? (The object's velocity is only zero for a instant. It does not permanently stop.)

2) How far from its starting (t = 0) position is the object when its velocity is zero for this instance?

3) At what time does the object again pass through the starting (x = 0) position?

4) If the object had initially been moving twice as fast (96 m/s), how far would it have gone before its velocity reached zero?

5) Suppose a second object begins moving with a constant speed of v = 48 m/s in the same direction from the same location at the same time as the object in part (4) above. At what time do the paths of these two objects once again cross?

Answer 1

Hi AA, here's how to tackle these types of problems:

(1) We need the equation which relates v_t to v₀, accel, and time: v_t = v₀ + at

Now, just plug in the given numbers: 0 = 48 + (-16) t

=> t = 3 seconds

(2) Now, we need the equation which relates distance to a and t: s_t = v₀t + (1/2) at²

= 48 x 3 + (1/2) (-16) x 9

= 144 - 72

= 72 meters from the start

(3) Now, we use the same equation, with s_t = 0: 0 = 48t +(1/2) (-16) t²

0 = 48t - 8t²

t² - 6t = 0

t(t - 6) = 0

=> t = 0 or t = 6

Since t = 0 corresponds to the initial position, the required answer is t = 6 seconds

(4) This is just parts (1) and (2) with a different v_o. I leave it to you as an exercise.

(5) Here, we want the time, t. when the distance the first object has traveled the same distance as the second object.

At time, t, the first object travels: s_t = 96t + (1/2) (-16) t²

At time, t, the second object travels 48t meters.

When their paths cross, these 2 distances are the same: => 96t - 8t² = 48t

8t² - 48t = 0

8t(t - 6) = 0

t = 0 or t = 6

t = 0 corresponds to the initial state, so the required answer is t = 6 seconds.

Answer 2

(1) v = u + at, you know v = 48, u = 0, a = -16. so solve for t

(2) s = ut + (1/2) at². You know u, a, and t from (1), so solve for s

(3) You have to find t when s=0. (you have a quadratic )

(4) repeat (1) and (2) with u = 96 (maybe you can find a quicker way !)

(5) Write an equation for s1 = .......................... the position of the first object.

Then write an equation for s2 = .................... the position of the 2nd object

Then find t when s1 = s2

Mike