Physics question?

Question

A speeder passes a parked cop car at a constant speed of 26.2 m/s.
At that moment, the cop starts from rest with a uniform acceleration of 2.7 m/s^2.
How much time passes before the speeder is overtaken by the cop?

Answer 1

Use the equation for distance covered. When the distance covered by both objects (driver and cop) is equal, it means the cop has caught up with the driver.

d_cop = v_0 * t + .5*a*t^2
in our case v_0 = 0 since the cop is at a stop.
a = 2.7

d_driver = v_d*t
(this needs no explanation)
v_d = 26.2

we need to find

d_cop = d_driver

so

.5*2.7*t^2 = 26.2*t

Now we solve for t, simple algebra

.5*2.7*t^2 - 26.2*t = 0

t * (1.35*t - 26.2) = 0

we don't care when t = 0 because that's when the driver passes the cop. so

(1.35*t - 26.2) = 0
t = 26.2 / 1.35
t = 19.4074 seconds

It takes the cop 19.41 seconds.

Answer 2

you need the time when their positions are equal taken that the origin is the point where the car starts

so let the position of the speeder be s and the position of car be x, then

s= t*v = t * 26.2
x= v(initial) + v(initial)*t + 0.5 a*t^2 =
0 + 0 + 0.5*2.7*t ^2= 1.35*t^2

now s=x implies

26.2*t=1.35*t^2 divide both sides by t

26.2=1.35t

t=26.2/1.35 = 19.41s

Answer 3

dont no

usee vectorss