49d
This case is about uniformly accelerated motion.
Given:
The initial speed was v takes distance d to stop after the brakes are applied.
Question:
What is the stopping distance if the car is initially traveling at speed 7.0v?
Assume that the acceleration due to the braking is the same in both cases. Express your answer using two significant figures.
The Process:
The list of variables to be considered is as follows.
The formula we follow for this problem are as follows:
[tex]\boxed{ \ v^2 = u^2 + 2ad \ }[/tex]
Step-1
We substitute v as the initial speed, distance of d, and zero for final speed into the formula.
[tex]\boxed{ \ 0 = v^2 + 2ad \ }[/tex]
[tex]\boxed{ \ v^2 = -2ad \ }[/tex]
Both sides are divided by -2d, we get [tex]\boxed{ \ a = \Big( -\frac{v^2}{2d} \Big) \ . . . \ (Equation-1) \ }[/tex]
Step-2
We substitute 7.0v as the initial speed, zero for final speed, and Equation-1 into the formula.
[tex]\boxed{ \ 0 = (7.0v)^2 + 2 \Big( -\frac{v^2}{2d} \Big)d' \ }[/tex]
Here d' is the stopping distance that we want to look for.
[tex]\boxed{ \ 2 \Big( \frac{v^2}{2d} \Big)d' = (7.0v)^2 \ }[/tex]
We crossed out 2 in above and below.
[tex]\boxed{ \ \Big( \frac{v^2}{d} \Big)d' = 49.0v^2 \ }[/tex]
We multiply both sides by d.
[tex]\boxed{ \ v^2 d' = 49.0v^2 d \ }[/tex]
We crossed out v^2 on both sides.
[tex]\boxed{\boxed{ \ d' = 49.0d \ }}[/tex]
Hence, by using two significant figures, the stopping distance if the car is initially traveling at speed 7.0v is 49d.
Keywords: a car traveling at speed v, takes distance d to stop after the brakes are applied, the stopping distance, if the car is initially traveling at speed 7.0v, the acceleration due to the braking is the same, two significant figures.
