How much time did it take for Batman to catch up to the criminal?

Question

In a daring car chase, the Batmobile and a Ford Focus start from rest at the same instant, with the Batmobile a distance behind the focus. The criminal has a constant acceleration of 3.10 m/s2 , and the Batman has an acceleration of 4.00 m/s2 . The Batmobile overtakes the Focus at a distance 50m from the criminal’s starting point.

a. How much time did it take for Batman to catch up to the criminal?

b. How large was the criminal’s lead?

c. How fast is the Batman going when he catches up to the focus?

Answer 1

Let

a = 3.1 m/s² be the criminal's acceleration,

A = 4.0 m/s² be Batman's acceleration,

s = 50 m be the distance the criminal travelled

S (unknown) be the distance Batman travelled,

t (unknown) be the time duration of the chase.

(a)

Since the criminal started from rest,

s = at²/2

From that

t = √(2s/a) = √(2 × 50 m / 3.1 m/s²) = 5.7 s

(b)

During the same time period t Batman travelled

S = At²/2

We could plug into it the just computed value of t, but we can get more accurate

calculation by going back to the definition of t.

Namely, the criminal's lead is

S - s =

At²/2 - at²/2 =

(A-a)t²/2 =

(A-a)((2s/a)/2) =

(A-a)(s/a) =

s(A-a)/a =

50×(4.0-3.1)/3.1 = 14.5 m

In other words, the Batman's lead (S-s) is the distance (s) travelled by the criminal

multiplied by the relative difference in Batman's acceleration (A-a)/a.

(c)

Batman's speed after accelerating for t seconds is

V = At = 4.0 m/s² × 5.7 s = 22.8 m/s