Question #0b63a

2 Answers
Jul 1, 2015

The cars should be at least 23m apart when the first one starts decelerating.


To find the distance the cars must be apart, we'll first look at how far the first car takes to come to a stop. (Your question didn't say the cars come to a stop, but I think it's a safe assumption)

Conversion of speeds first so all of the units will cancel.
#90"km"/"hr" = 90000"m"/"hr" = 90000"m"/"3600s"=25m/s#

Car 1
#0^2 = (25"m"/"s")^2+2(-7.5"m"/"s"^2)d#
#d = 41.67m#

We'll repeat this for Car 2 which has two distances to determine

Car 2
Distance Traveled before deceleration started
#25"m"/s = "d"/"0.5s" => d = 12.5m#

Distance traveled while decelerating
#0^2 = (25"m"/"s")^2+2(-6"m"/"s"^2)d#
#d = 52.08m#

Total distance Car 2 travels is #64.58m#

So if Car 1 travels 41.67m and Car 2 travels 64.58m before stopping, the difference in distances must be our minimum starting distance, which is approximately 23m (22.91m if you want to be exact).

Jul 1, 2015

Here's how you can approach this problem.



When a balloon that travels directly upwards is 20m high from the ground, its velocity is #"5 m s"^(-1)#. At this moment an object is released downward from it.

  • After 1s and 2s, what are the positions (from the ground) and the velocities of the object?
  • After how long will the object hit the ground? And with what speed?

The idea behind this problem is that, at the time of its release, the object will have an initial velocity equal to the velocity of the balloon.

Moreover, this initial velocity will be pointed upwards.

So, let's assume that upwards is the positive direction and downwards is the negative direction.

So, once the object is released, it will travel upwards with an initil velocity of #v_0 = "5 m/s"#. As you would image, it is going to travel upwards for a while, then stop at the top of its movement, i.e. when it reaches maximum height, then begin to drop towards the ground.

After 1 second, its position will be

#y_1 = v_0 * t - 1/2 * g * t^2#

#y_1 = 5"m"/cancel("s") * 1cancel("s") -1/2 * 9.8 "m"/cancel("s"^2) * 1^2cancel("s"^2) = "0.1 m"#

Since the displacement is positive, the object is actually 0.1 m above the point of its release. Therefore, the distance from the ground after 1 second will be

#d_1 = 20 + 0.1 = color(green)("20.1 m")#

Its velocity will be

#v_f = v_0 - g * t#

#v_1 = 5"m"/"s" - 9.8"m"/"s"^(cancel(2)) * 1cancel("s") = color(green)(-"4.8 m/s")#

This means that the object is already on its way down and is actually heading towards the ground.

After 2 seconds, you'll get

#y_2 = 5"m"/cancel("s") * 2cancel("s") -1/2 * 9.8 "m"/cancel("s"^2) * 2^2cancel("s"^2) = -"9.6 m"#

This means that the object is below its release point, so the distance from the ground will be

#d_2 = 20 - 9.6 = color(green)("10.4 m")#

Its velocity will be

#v_2 = 5"m"/"s" - 9.8"m"/"s"^(cancel(2)) * 2cancel("s") = color(green)(-"14.6 m/s")#

To get the time of impact with the ground, use the object's vertical displacement. Since it started at 20 m above the ground, and ends up on the ground, it's vertical displacement will be

#y = -"20 m"#

Therefore, the time it takes for the object to reach ground level will be

#y = v_0 * t - 1/2 * g * t^2#

#-20 = 5 * t - 4.9 * t^2 => -4.9t^2 + 5t + 20 = 0#

This quadratic equation will prodcue two solutions. Pick the positive one to get

#t = color(green)("2.59 s")#

The velocity upon impact will be

#v_"impact" = v_0 - g * t#

#v_"impact" = 5"m"/"s" - 9.8"m"/"s"^(cancel(2)) * 2.59cancel("s") = color(green)(-"20.4 m/s")#