A student is running at her top speed of 4.6 m/s to catch a bus, which is stopped at the bus stop. When the student is still a distance 38.9 m from the bus, it starts to pull away, moving with a constant acceleration of 0.177 m/s^2.
A) For how much time does the student have to run at 4.6 m/s before she overtakes the bus?
B) For what distance does the student have to run at 4.6 m/s before she overtakes the bus?
C) When she reaches the bus, how fast is the bus traveling?
D) What is the minimum speed the student must have to just catch up with the bus?
E) For what time does she have to run in that case?
F) For what distance does she have to run in that case?
distance the student runs=38.9+4.6t
distance the bus goes=1/2 a t^2
in the same time, the respective distances are equal. Set them equal, and solve for time.
Can I use the same equations to solve for the distance in part B?
To solve these problems, we need to use kinematic equations that relate variables such as distance, speed, time, and acceleration. Let's go through each question step by step.
A) For how much time does the student have to run at 4.6 m/s before she overtakes the bus?
To find the time, we can use the equation: distance = speed × time. Rearranging the equation, we have:
distance = speed × time
time = distance / speed
Plugging in the values, we have:
time = 38.9 m / 4.6 m/s
Calculating this, we get:
time = 8.46 seconds
Therefore, the student has to run for approximately 8.46 seconds before she overtakes the bus.
B) For what distance does the student have to run at 4.6 m/s before she overtakes the bus?
We can use the same equation as in part A to find the distance. Rearranging the equation, we have:
distance = speed × time
Plugging in the values, we have:
distance = 4.6 m/s × 8.46 seconds
Calculating this, we get:
distance = 38.76 meters
Therefore, the student has to run approximately 38.76 meters before she overtakes the bus.
C) When she reaches the bus, how fast is the bus traveling?
To find the speed of the bus, we need to use the equation of motion that relates final velocity, initial velocity, acceleration, and time:
final velocity = initial velocity + (acceleration × time)
In this case, the bus starts from rest (initial velocity = 0), and the student has been running for 8.46 seconds at a constant speed of 4.6 m/s. The acceleration of the bus is given as 0.177 m/s².
Plugging in the values, we have:
final velocity = 0 + (0.177 m/s² × 8.46 seconds)
Calculating this, we get:
final velocity = 1.4962 m/s
Therefore, when the student reaches the bus, the bus is traveling approximately 1.4962 m/s.
D) What is the minimum speed the student must have to just catch up with the bus?
To determine the minimum speed needed by the student to just catch up with the bus, we need to consider the relative motion between the student and the bus.
Since the student is initially 38.9 meters away from the bus and starts running at her top speed of 4.6 m/s, the minimum speed required by the student is the same as the speed of the bus when the student catches up.
Therefore, the minimum speed the student must have to just catch up with the bus is 1.4962 m/s (as calculated in part C).
E) For what time does she have to run in that case?
In this case, the time required for the student to catch up to the bus is the same as the time taken by the bus to cover the initial distance of 38.9 meters. We can use the equation of motion:
distance = initial velocity × time + (0.5 × acceleration × time^2)
Since the bus starts from rest (initial velocity = 0) and the acceleration of the bus is given as 0.177 m/s², we can solve for time using the above equation.
Plugging in the values, we have:
38.9 meters = 0 × time + (0.5 × 0.177 m/s² × time^2)
This equation is a quadratic equation. Solving it, we find two possible solutions: time = 0 seconds (not applicable in this context) and time = 15.169 seconds.
Therefore, in this case, the student has to run for approximately 15.169 seconds.
F) For what distance does she have to run in that case?
In this case, the distance covered by the student before she catches up to the bus is the same as the initial distance between them, which is 38.9 meters (as given in the problem). Hence, the student has to run for approximately 38.9 meters in this case.