Suvat equations - car braking hard to avoid a collision

[Apothem]

Suvat equations -- car braking hard to avoid a collision

Homework Statement

A car (A) is speeding along a rural road, of width 4m with a speed of 25m/s when the driver sees another vehicle (B) of length 3.5m just starting to cross the road at a point 40m ahead. The drivers reaction time is 0.80s, and the maximum deceleration with the brakes fully applied is 5.5m/s/s, If the car can be assumed to decelerate uniformly without swerving etc., calculate:

a) The distance traveled by the car during the drivers reaction time
b) The car's (A) velocity when it reaches the position of the other vehicle (B)
c) Total time which has elapsed from first sighting until car A reaches vehicle B
d) The minimum constant velocity of vehicle B so that the car A does not collide with it

Homework Equations

SUVAT Equations

The Attempt at a Solution

a)
s=
u=25m/s
v=
a=-5.5m/s/s
t=0.80

s=ut+0.5(at²)
s=18.24m

b)
s=40m
u=25m/s
v=
a=-5.5m/s/s
t=

v²=u²+2as
v²=25²+2(-5.5)(40)
v=13.60m/s

c)
s=40m
u=25m/s
v=13.60m/s
a=-5.5m/s/s
t=

v=u+at
13.60=25-5.5t
t=(13.60-25)/-5.5=2.07s

d) Rather unsure on this one:
s= 3.5m
u=
v=
a=0m/s/s
t=2.07s

I never calculated anything for d) because I am really unsure.

Any help is appreciated!

 

[Tanya Sharma]

a)
s=
u=25m/s
v=
a=-5.5m/s/s
t=0.80

s=ut+0.5(at²)
s=18.24m

Reaction time is the time taken by the driver to start applying brakes on seeing the obstacle(car B in this case) .During the reaction time the car A is traveling at 25m/s with no deceleration.

b)
s=40m
u=25m/s
v=
a=-5.5m/s/s
t=

v²=u²+2as
v²=25²+2(-5.5)(40)
v=13.60m/s

The item in red should be 20m

 

[Apothem]

During the reaction time the car A is traveling at 25m/s with no deceleration.

Ok therefore:

a)
s=
u=25m/s
v=
a=0m/s/s
t=0.80

s=ut+0.5(at²)
s=25(0.80)+0
s=20m

Subsequently for b)

b)
s=40-20=20m
u=25m/s
v=
a=-5.5m/s/s
t=

v²=u²+2as
v²=25²+2(-5.5)(20)
v=20.1m/s

And for c)
c)
s=20m
u=25m/s
v=20.1m/s
a=-5.5m/s/s
t=

v=u+at
20.1-25/-5.5=t=0.89s
So total t=0.89+0.80=1.69s

Is this right?

 

[Tanya Sharma]

Ok therefore:

a)
s=
u=25m/s
v=
a=0m/s/s
t=0.80

s=ut+0.5(at²)
s=25(0.80)+0
s=20m

Subsequently for b)

b)
s=40-20=20m
u=25m/s
v=
a=-5.5m/s/s
t=

v²=u²+2as
v²=25²+2(-5.5)(20)
v=20.1m/s

And for c)
c)
s=20m
u=25m/s
v=20.1m/s
a=-5.5m/s/s
t=

v=u+at
20.1-25/-5.5=t=0.89s
So total t=0.89+0.80=1.69s

Is this right?

Yes...that's right

 

[Apothem]

Yes...that's right

Now comes part d)

My thinking is:
s=3.5m
u=
v=
a=0m/s/s
t=1.69s

Then work out u?

 

[Tanya Sharma]

Now comes part d)

My thinking is:
s=3.5m
u=
v=
a=0m/s/s
t=1.69s

Then work out u?

No...

How much distance car B needs to travel to completely cross the road ?

 

[Apothem]

No...

How much distance car B needs to travel to completely cross the road ?

4 metres

 

[adjacent]

4 metres

Nope.The car has to COMPLETELY cross the road.The width of the road is 4m.Length of the car is 3.5m
Think about it.

 

[Apothem]

Nope.The car has to COMPLETELY cross the road.The width of the road is 4m.Length of the car is 3.5m
Think about it.

7.5 metres?

 

[Tanya Sharma]

4 metres

Car is not a particle .It has a length .The distance to be covered is the width of the road + length of the car .

 

[Tanya Sharma]

7.5 metres?

Yes...

 

[Apothem]

So am I right to say:
d)
s=7.5m
u=
v=
a=0m/s/s
t=1.69s

s=ut+0.5(at^2)
50=1.69u
u=50/1.69=29.6m/s

 

[Tanya Sharma]

So am I right to say:
d)
s=7.5m
u=
v=
a=0m/s/s
t=1.69s

s=ut+0.5(at^2)
50=1.69u
u=50/1.69=29.6m/s

s should be 7.5 not 50 .

 

[Apothem]

s should be 7.5 not 50 .

Sorry I was looking at my maths homework just before hand, and 50 was the answer to a question, wasn't paying attention :shy: