A physics student who is late for school is trying to catch her bus. The student is 10 m behind the bus as the bus starts from rest and begins to accelerate at 0.5 m/s2 away from her. If the student continues running at a steady 3 m/s, will she catch her bus?
distancestudent=3*time + 10
distancebus=1/2 .5 *time^2
set them equal, if she catches the bus..
3t+10=t^2/4
t^2-12t+40=0
t=(12+-sqrt(144-160))/2 notice the sqrt of a negative term...means the solution cannot be real, she does not catch the bus, except in imaginary time.
d bus = .5 a t^2 = .25 t^2
d student = d bus + 10 = 3 t
so
.25 t^2 = 3 t - 10
.25 t^2 -3 t + 10 = 0
t^2 - 12 t + 40 = 0
t = [ 12 +/- sqrt(144 -160) ] /2
oh my, complex t
student only catches bus in imagination
There is a small error. It should be -10, not +10, in the original equation of d=3t-10
To determine if the student will catch the bus, we can calculate the time it takes for the student to reach the same position as the bus.
We can use the kinematic equation:
\[ \Delta x = v_i t + \frac{1}{2} a t^2 \]
where:
- \(\Delta x\) is the displacement or distance between the student and the bus (10 m),
- \(v_i\) is the initial velocity of the bus (0 m/s),
- \(a\) is the acceleration of the bus (0.5 m/s\(^2\)),
- \(t\) is the time taken.
Since both the student and the bus move in the same direction, we need to set their displacements as equal:
\[ 10 = (0 \times t) + \frac{1}{2} (0.5) t^2 \]
Simplifying the equation:
\[ 10 = \frac{1}{2} (0.5) t^2 \]
\[ 10 = 0.25t^2 \]
To solve for \(t\), we can multiply both sides of the equation by 4 and then take the square root:
\[ 40 = t^2 \]
\[ t = \sqrt{40} \]
\[ t \approx 6.32 \]
The time it takes for the student to catch the bus is approximately 6.32 seconds.
Now, we can determine the distance the student covers in 6.32 seconds:
\[ \text{Distance} = \text{speed} \times \text{time} \]
\[ \text{Distance} = 3 \times 6.32 \]
\[ \text{Distance} \approx 18.96 \]
Therefore, the student will not catch the bus as the distance covered by the student (approximately 18.96 m) is less than the initial 10 m distance between them.