Question

A car traveling at 32 m/s passes a policeman wo immediately accelerates at 2.43 m/s^2 from rest in pursuit. To the nearest second how long does it take the policeman to catch the speeder?

Answer 1

Well do you know the answer of this question ?

Answer 2

I don't, and I'm trying to see if any of the formulas that you gave would work here.

Answer 3

Well it's a tricky one !
You'll have to use one of the formulas

Answer 4

Well, we have acceleration (a) for two cars...

Answer 5

Just imagine the situation, as soon as the policemen sees the car, he starts following it and finally catches it. Here, distance covered and time taken by the car and the policemen during the pursuit will be same

Answer 6

The acceleration given is just for the policemen's car [if he is chasing by a car :p]. The other car is moving with a constant velocity of 32 m/s

Answer 7

Confusing, lol.

Answer 8

Hmm, would we have to use the first one?

Answer 9

v = u + at ?

Answer 10

And solve ti for t?

Answer 11

Let t be the time taken which would be same for both. Now,

\(\sf 32 \times t=0\times t+2.43\times t^2\)

Now solve it for t

Answer 12

?

Answer 13

Still trying to figure it out. It's onl proving that I need to review basic Algebra...

Answer 14

Let t be the time,
then distance traveled by first car = 32*t [in uniform motion distance=speed*time]
distance traveled by police = 0*t+1/2*2.43*\(\sf t^2\)
Since distance covered during the pursuit by both the cars would be same we can equate the two equations and write it as


\(\sf 32 \times t=0\times t+2.43\times t^2\)

Solving it for t we get

32/2.43=t
t=13.16 s

Answer 15

is it clear @MoonlitFate ?

Answer 16

Yeah, I didn't notice that the equation was a quadratic. :( So, I was making things so much more difficult than they needed to be.