A personne walking at a speed of 1.7m/s see's a stoped bus 25m from him. At this moment this person starts to run with an acceleration of 1,3m/s2. After 4 secondes this personne has reached his maximum speed et it stays constant, but at the same time the bus starts moving and accelerates 2,6m/s2.
Does this person suceed in catching the bus? If so, at what time and at what distance, if not what was the smallest distance between the person and the bus.
- Im sorry if there any mistakes, i translated from french.
Ive got this so far:
Alright well, by using the initiale speed of 1,7m/s and 4s, i found his constante speed of 6.9m/s. So after that i used the acceleration and his speed to determine the distance he travelled, 17,2m. i subtracted 17,2m from 25m, which left 7,8m between the bus and the person. Then im stuck, how do i incorporate both speeds in one equations to see if he caught the bus or not?
The have to get to the same point, which is 25m further for the person.
distance= 25+ 1.7*time + 1/2 1.3 * 4^2 + max speed*(time-4)
max speed= 1.7 + 1.3*4 so put that in the first equation for max speed.
and distance= 1/2 *2.6* (t-4)^2
set these two distances equal, then solve for time. If time is real, and positive, the person catches the bus.
To solve this problem, you already have the initial speed of the person and the acceleration. You correctly found the constant speed of the person to be 6.9 m/s after 4 seconds.
Next, let's set up an equation to find the distance covered by the person. The distance covered in this scenario can be broken down into three parts:
1. The distance covered during the initial 4 seconds, at a constant acceleration of 1.3m/s^2:
distance1 = (1/2) * acceleration * time^2
= (1/2) * 1.3 * 4^2
= 10.4m
2. The distance covered after reaching the constant speed, at the maximum speed of 6.9m/s:
distance2 = max speed * (total time - 4s)
= 6.9 * (t - 4)
3. The initial distance from the person to the bus is 25m, so the total distance covered by the person is:
total distance = distance1 + distance2 + initial distance
= 10.4 + (6.9 * (t - 4)) + 25
= 6.9t - 13.6
Now, let's consider the distance covered by the moving bus. The distance covered by the bus can be expressed as:
distance bus = (1/2) * acceleration bus * (t - 4)^2
= (1/2) * 2.6 * (t - 4)^2
= 1.3 * (t - 4)^2
Since the person catches the bus when their distances are equal, we can set the two equations equal to each other and solve for time:
6.9t - 13.6 = 1.3 * (t - 4)^2
Simplifying this equation will give you a quadratic equation that can be solved to find the time at which the person catches the bus. If the time is real and positive, then the person catches the bus. If not, then the person does not catch the bus.
Note: Make sure to use the correct units for time, acceleration, and velocity when solving this problem.