Question

A car moving along a straight highway with speed
144km/h is brought to stop within a distance of 400m
with uniform retardation. How long does it take for the
car to stop?​

Answer 1

Answer:

Explanation:

Initial velocity of the car (u) = 126km/hr = 35m/s

Distance travelled before coming to rest (s)=200m

Final velocity of the car (v) =0 (as it comes to rest)

Using the 3rd equation of motion,

v^2 - u^2 = 2as

a = (v^2 - u^2)/2s

=(0 - 35^2)/2×200

= –1225/400

= –3.06m/s^2

Let time taken to come to rest is t sec

Using the 1st equation of motion,

v = u + at

0 = 35 – 3.06×t

t =35/3.06

=11.43 sec

HOPE IT HELPS..... :-)

Answer 2

Answer:

5.5=a

Explanation:

v=0

u=126khm/h=144*18/5=40m/s

s=144m

a=?

v^2= u^2 +2as

o=1600+2*a*144

1600+a288=0

a288=-1600

a=1600/288

a=5.5