Question

A car is moving at rate of 72km/h and applies brake which provide a retardation of 5ms-2. 1. How much time does ithe car take to stop. 2.how much distance does the car cover before coming to rest? 3. what would be the stopping distence needed if speed of the car is doubled?

Answer 1

To answer this question use the formula:
v=u+at
s={(u+v)/2}t
v2=u2+2as
s=ut+1/2at2  where s-displacement, v-final velocity, u-initial velocity, a-acceleration/retardation, t-time
i)How much time does the car take to stop?

in this case V=72km/hr (convert to m/s=20m/s), u=0m/s, a=m/s2
since v=u+at...at=v-u....t=(v-u)/a
therefore t=(20-0)/5.....=20/5 =4s
therefore it takes 4 seconds to stop

ii)distance covered before stopping: displacement (s)

s= [(u+v)/2]t....=0+20/2 x 4....= 10 x 4= 40ms
hence distance covered is 40 meters

iii)what is stopping distance if speed is doubled?

that means v= 40m/s
hence displacement s,
s= [(u+v)/2]t....= 40/2 x 4
    displacement therefore at double speed =  80 meters

Answer 2

Answer:

u = 72 km/h

= 72×5/18

= 20 m/s

a = -5 m/s²

(i) v = 0

v = u+at

0 = 20-5t

-5t=-20

t = 20/5

t = 4 s

(ii) s = ut+1/2at²

= 20×4 + 1/2×-5×4²

= 80 - 1/2×5×16

= 80 - 40

= 40 m

(iii) Speed is doubled :-

u = 20×2 = 40 m/s

v = 0 m/s

a = -5 m/s²

v²-u²=2as

0²-40²=2×-5s

-1600=-10s

s = -1600/-10

s = 160 m