Question

In a drag race, the position of a car as a function of time is given by x=bt2, with b = 2.045m/s2 . In an attempt to determine the car's velocity midway down a 400-m track, two observers stand at the 165-m and 235-m marks and note the time when the car passes.What value do the two observers compute for the car's velocity over this 70m stretch?

Answer 1

The Position of car is given by X= bt²     (It means inital velocity of car u is 0 )

At first we calculate Time taken to reach 165 m mark.

165=2.045t²

t=8.98 s

velocity at 165 m mark can be calculated by

V=u+at

(here u=0 , a=2.045m/s² , t=8.98 s)

V=2.0458.98

V=18.36 m/s

Now taken to reach the 235 m mark

t=10.72 s

and velocity at 235 m mark is

V=u+at

(u=0 ,t=10.72 s,a=2.045m/s²)

V=2.04510.72

V=21.92 m/s

Now time taken to pass 70 m stretch t'=(10.72 - 8.98)

                                                    t'=1.74 s

observer will compute average velocity V_avg between two given point

V_avg=Total distance/total time

V_avg = 70/1.74

    V_avg    =40.23 m/s   (ANS)