Question

A student is running at her top speed of 5m/s to catch a bus, which is stopped at a bus stop. When the student is still 80m behind the bus, it starts to pull away, moving with a constant acceleration of 0.1m/s2. (a) How much time will be needed for student to catch the bus? (b) What distance does the student have to run before she catches the bus? (c) When she reaches the bus, how fast is the bus traveling? (d) If the student is running too slow, she will not catch the bus. What is the minimum velocity vm with which she should run and still catch the bus? (e) At what time she would catch the bus now?

Answer 1

(a) Let the student catches the bus at t seconds.

Distance d₁ travelled by student after t seconds , if she is running at a speed 5 m/s is given by ,

d₁ = ( 5 t )

Distance d₂ travelled by bus in t seconds = (1/2) a t² = (1/2) 0.1 t² = 0.05 t²

where a is acceleration

Distance between student and bus before bus started = 80 m

Hence we have , d₁ = d₂ + 80

5t = .05 t² + 80 or   t² -100 t + 1600 = 0

(t-20)(t-80) = 0 , hence t = 20 s or t = 80 s

( at t = 20 s, speed of student is greater than speed of bus. Suppose the student is not boarding the bus but runs ahead of bus, then she will overtake the bus. Speed of bus keep on increasing . Hence at 80 s , bus will catch the student. This is the reason we get two solutions for quadratice equation. We will take only the first solution t = 20 s )

Conclusion :- Student will catch the bus after 20 s

---------------------------------------------

(b) Distance run by student for 20s time duration = speed time = 5 20 = 100 m

-----------------------------------------

(c) Speed of bus after 20 s , v = a t = 0.1 20 = 2 m/s

-----------------------------------------------

(d) if spped of student is u then distance covered by student in t seconds is ( u t )

distance covered vy bus in t seconds = (1/2) 0.1 t² = 0.05 t²

Condition for catching the bus :- ( u t ) = 0.05 t² + 80

Above equation can be simplified as, t² - (20 u ) t + 1600 = 0 .................(1)

above quadratic equation has real solutiuon , only if or

Hence minimum speed u = 4 m/s

--------------------------------------------------------

(e) For minimum speed u = 4 m/s , quadratic equation (1) can be written as

t² - 80 t + 1600 = 0 or (t-40)² = 0

Hence t = 40 s

With minimum speed u = 4 m/s , student will catch the bus after 40 s