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A student is standing 50 m behind a bus. As soon as the bus begins its motion with acceleration 1 ms -2 , the student starts running towards the bus wit a constant velocity u. Assuming the motion to be along a staright road, what is the minimum value of u, so that the student is able to catch the bus? 5 ms -1 8 ms -1 10 ms -1 12 ms -1 The answer is C. 10 ms -1

A student is standing 50 m behind a bus. As soon as the bus begins its motion with acceleration 1 ms-2, the student starts running towards the bus wit a constant velocity u. Assuming the motion to be along a staright road, what is the minimum value of u, so that the student is able to catch the bus?
  1. 5 ms-1
  2. 8 ms-1
  3. 10 ms-1
  4. 12 ms-1
The answer is C. 10 ms-1
 

Grade:11

3 Answers

Vikas TU
14149 Points
7 years ago
To catch the bus, their distance covered should be same in same time.
For bus let it covered x distance from initial position,
therefore,
x = 0 +  0.5*1*t^2
x = t^2/2 ….............................................(1)
 
For student,
(x + 50) = u*t........................................(2)
substituute eq. (1) from eq. (2)
t^2/2 + 50 = ut
or
u = t/2 + 50/t
du/dt = ½ – 50/t^2 = 0
50/t^2 = 0.5
t^2 = 50/0.5 = 100
t = 10 seconds
 
u = 10/2 + 50/10
  = 5 + 5
  = 10 m/s
Sandip
11 Points
5 years ago
Let the time required to catch the bus = t
Now, distance travelled by the bus in “t” s =  1/2t^2  [ since initial velocity of bus =0 & a= 1m/s^2]
Distance travelled by the student in time “ t ” s = ut
therefore      ut = 1/2t^2 + 50 …...... Eqn (i)
now the velocity “ v “of the bus after “ t “ s      v = t, …..... Eqn (ii)          [ Since v= u + at, and u = 0 and a = 1m/s^2]
 
now the relative velocity of the student and the bus at the time of catching will be =0, ie u = v
therefore u = t  [v = t .. eqn (ii) & u = v]
now putting the value  of u = t  in Eqn (i) we get 
t^2 = 1/2t^2 + 50 = (t^2 + 100)/2
or 2t^2 = t^2 + 100
or t^2 =  100 
or t = 10
so u = 10m/s   [ u = t]
 
 
 
 
AdyGreg
15 Points
5 years ago
Modified answerTo catch the bus, their distance covered should be same in same time.
For bus let it covered x distance from initial position,
therefore,
x = ut + 1/2 at2
x = 0 + 0.5x1xt2
x = t2/2 ….............................................(1)
 
For student,
(x + 50) = u x t........................................(2)
substituute eq. (1) from eq. (2)
t2/2 + 50 = ut
or
u = t/2 + 50/t
du/dt = 1/2 – 50/t2 = 0
50/t2 = 0.5
t2 = 50/0.5 = 100
t = 10 seconds
 
u = 10/2 + 50/10
  = 5 + 5
  = 10 m/s

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