How to solve Motion Question?

Question

A policeman (on a motor cycle, with the engine going) is hiding behind a bush on the side of the road when a speeding truck passes with a speed of #96.5 kmh^-1#. As the truck passes he sets off with an acceleration of #1.83ms^-2#, which he maintains until he overtakes it.

Calculate:
(a) the time he took to overtake the truck
(b) the distance he travelled in so doing
(c) the velocity, in #kmh^-1#, of the motor cycle when it catches the truck.

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Answer 1 · 2018-02-25T01:24:49

I tried this:

Have a look:

enter image source here

so:
#d=26.8xxcolor(red)(29)=777m#

and from:
#v_f=v_1+at_1#
#v_f=1.83xx color(red)(29)=53m/s=190km/h#

Answer 2 · 2018-02-25T17:35:14

(a) 29.3 s
(b) 785 m or 0.785 km
(c) 53.6 m/s or 202 km/hr

(a)

The key to this problem is the fact that the two vehicles travel the same distance in the same time.

First we need to convert the speed of the truck into m/s:

#sf(v=(96.5xx1000)/(60xx60)=26.8color(white)(x)"m/s")#

The speed of the truck is constant and the distance is the same so we can write:

#sf(s=vt)# for the truck.

and #sf(s=cancel(ut)+1/2at^2)# for the police bike.

Putting these equal:

#sf(vcancel(t)=1/2at^cancel(2))#

#:.##sf(t=(2v)/a=(2xx26.8)/(1.83)=29.3color(white)(x)s)#

(b)

#sf(s=vt)#

#:.##sf(s=26.8xx29.3=785color(white)(x)m=0.785color(white)(x)"km")#

(c)

We can use:

#sf(v=u+at)#

#:.##sf(v=0+1.83xx29.3=53.6color(white)(x)"m/s")#

#sf(v=(53.62xx60xx60)/(1000)=202color(white)(x)"km/hr")#

(This is about 126 mph)