When Will Tom Overtake?
Tom the cat, on seeing Jerry the mouse at distance
d
=
5
m
, starts running towards Jerry with velocity
5
m/s
and acceleration
2
.
5
m/s
2
in order to catch it. Simultaneously, Jerry starts running (directly away from Tom) from rest with acceleration
a
m/s
2
.
To 2 decimal places, what is the maximum value of a , for which Tom will be able to catch Jerry?
Image credit: Wikipedia
The answer is 5.00.
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7 solutions
Write equations of motion get a quadratic and discriminant should be greater than zero.
ya 5 is correct
of course it should be. but if it is greater than 5, then tom couldn't catch jerry.
why should the discriminant be greater than zero?
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you can write equation for the case when tom catches jerry at time T. let position of tom be S t o m = 5 t + ( 1 / 2 ) ( 2 . 5 ) t 2 the position of jerry at time t will be S j e r r y = 5 + ( 1 / 2 ) A t 2
for S t o m = S j e r r y , we get equation
( A − 2 . 5 ) t 2 − 1 0 t + 1 0 = 0
for this quadratic equation to have real solution, D = 200 - 4A should be greater than equal to zero. that means for tom to catch jerry in real time, D should be greater than or equal to zero i.e. A should be less than equal to 5 or there will be no "real" solutions to the equations.
hence the equation will have no real solution for A>5 and hence jerry can avoid being caught if A>5
We get the quadratic equation in variable 't'(where 't' denotes time), and because time is always a real value,that's why the discriminant ought to be zero
Let the time in which tom catches jerry be x sec.
tom should travel 5 m more than jerry.
⇒ 2 1 A x 2 + 5 = 5 x + 2 1 ∗ 2 . 5 x 2
⇒ A = 2 x 2 5 x 2 + 2 0 x − 2 0
To obtain maximum value of A we need to maximize 2 x 2 5 x 2 + 2 0 x − 2 0
which is 5 at x = 2 .
For this to happen the moment tom catches jerry, the elapsed time would e equal, tom would have traveled only 5m more than jerry, and their ending velocities will be equal...
VELOCOTY OF JERRY AT ANY TIME T IS (At) DISPLACEMENT OF JERY IS 1/2 At2 DISPLACEMENT OF TOM IS 5+ DISPLACEMENT OF JERRY AND VELOCITY OF TOM IS EQUAL TO THE VELOCITY OF JERRY
Or an Alternate way , Switch to tom's inertial reference frame , acceleration = 2.5-A, Initial velocity = 5 We have by equation of motion, v v + u u = 2(2.5-A)x5 So , v = sqrt(10(2.5-A)+5*5) Since sqrt is defined only for positive numbers solve the inequality an you get A <= 5.0
jerry itself is at distance 5m & tom starts running with a speed of 5m/s. h e instantly catches him. ha ha what a wierd question is this.
no. jerry is running too with A m/s2. if A is more than 5 m/s2, tom will be dead trying.
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then tom will run faster in order to catch jerry. he wont die so easily
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No he won't, there is spike and son in front of him. Tom will run the other way and catch tweety instead. Haha.
Let this the function of Tom's distance: f ( t ) = 1 . 2 5 t 2 + 5 t
And this the function of Jerry's distance: g ( t ) = 2 A t 2 + 5
Now, the moment when Tom catches Jerry will be: f ( t ) = g ( t )
1 . 2 5 t 2 + 5 t = 2 A t 2 + 5
Reorder and multiply by 4 :
( 2 A − 5 ) t 2 − 2 0 t + 2 0 = 0
Calculate the discrimimant:
δ t = 8 0 0 − 1 6 0 A
The maximum value of A will be when δ t = 0 , so A = 5