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1. 4The Highway Code states that the braking distance for a car travelling at 30 mph (13 m/s) is 14 m. Calculate the deceleration assumed by the Highway Code.

Plus, is decreasing acceleration with velocity as the y axis and time as the x, is decreasing acceleration a curved line ( like the top left quarter of a circle)
2. (Original post by Sraf)
4The Highway Code states that the braking distance for a car travelling at 30 mph (13 m/s) is 14 m. Calculate the deceleration assumed by the Highway Code.

Plus, is decreasing acceleration with velocity as the y axis and time as the x, is decreasing acceleration a curved line ( like the top left quarter of a circle)
you can do this with a SUVAT...
v2=u2+2as
since the car is decelerating to a halt, the acceleration will be negative and the final velocity is zero.

---
check whether it's velocity v time or distance v time graph - it'd be unusual to deal with non constant acceleration with students who are just beginning.
3. (Original post by Joinedup)
you can do this with a SUVAT...
v2=u2+2as
since the car is decelerating to a halt, the acceleration will be negative and the final velocity is zero.

---
check whether it's velocity v time or distance v time graph - it'd be unusual to deal with non constant acceleration with students who are just beginning.
Exact question is

4 The Highway Code states that the braking distance for a car travelling at 30 mph (13 m/s) is 14 m. Calculate the deceleration assumed by the Highway Code.

0.9 m/s 141 m/s 0.5m/s 6m/s
That's the options
4. (Original post by Joinedup)
you can do this with a SUVAT...
v2=u2+2as
since the car is decelerating to a halt, the acceleration will be negative and the final velocity is zero.

---
check whether it's velocity v time or distance v time graph - it'd be unusual to deal with non constant acceleration with students who are just beginning.
It's a velocity time graph btw
5. (Original post by Sraf)
It's a velocity time graph btw
well if it goes from a steep upward gradient to a less steep upward gradient over a period of time it means that the acceleration has decreased over that period of time.

are you able to rearrange v2=u2+2as
to find a?

v is final velocity - since the car slows to halt this is zero
u is initial velocity 13 m/s
s is 14m
a is the acceleration... and will be a negative number because deceleration is negative acceleration
6. so if you sketch a velocity-time graph you get a triangle with height 13 and base ?

the area of the triangle is the distance...

so 13 x ? /2 = 14

find ? and then work out the gradient of the hypotenuse.... this is the acceleration.

the units should be m/s2 not m/s

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