Question

A driver of car a travelling at 54km/h applies the brakes and stops the car in 4s. Another driver of car b travelling at 36km/h applies brakes and stops the car in 6s. Plot speed time graph for two cars. Which of the two cars travelled farther before stopping

Answer 1

For car A
U=54m/s^2
V=0
T=4s
Therefore acceleration=
By first equation of motion ( v=u-at) for retardation
0=54-4a
A=54/4=13.5m/s^2
Therefore distance travelled=
By second equation of motion s= ut-1/2 at^2( for retardation)
S= 54x4-1/2x13.5x16
S=216-108=108m
For car B
U=36m/s
V=0
T=6s
Therefore acceleration=
V= u-at
0=36-6a
a= 6m/s^2
Therefore distance travelled=
S=ut-1/2at^2
S=36x6-1/2x6x36
S= 216 - 108 =108 m
Therefore both the car travelled same distance i.e 108m before stopping

Answer 2

Answer:

For car A

U=54m/s^2

V=0

T=4s

Therefore acceleration=

By first equation of motion ( v=u-at) for retardation

0=54-4a

A=54/4=13.5m/s^2

Therefore distance travelled=

By second equation of motion s= ut-1/2 at^2( for retardation)

S= 54x4-1/2x13.5x16

S=216-108=108m

For car B

U=36m/s

V=0

T=6s

Therefore acceleration=

V= u-at

0=36-6a

a= 6m/s^2

Therefore distance travelled=

S=ut-1/2at^2

S=36x6-1/2x6x36

S= 216 - 108 =108 m

Therefore both the car travelled same distance i.e 108m before stopping

Explanation: