Question

1. In 1976, Kitty Hambleton of the United States drove a rocket-engine

car to a maximum speed of 965 km/h. Suppose Kitty started at rest
and underwent a constant acceleration with a magnitude of 4.0 m/s2.
What distance would she have had to travel in order to reach the maxi-
mum speed?

Answer 1

Kitty would have had to travel a distance of approximately 18,058.7 meters (or 18.06 kilometers) to reach her maximum speed of 965 km/h with a constant acceleration of 4.0 m/s^2.

Step-by-step explanation:

To find the distance Kitty had to travel to reach her maximum speed, we need to use the equation: v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance. In this case, Kitty started from rest (u = 0), her final velocity was 965 km/h (v = 965 km/h * 1000 m/km * 1 h/3600 s = 268.06 m/s), and her acceleration was 4.0 m/s^2. Plugging these values into the equation, we can solve for s:

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

s = (268.06^2 - 0^2) / (2 * 4.0)

s = 18058.7 m

Therefore, Kitty would have had to travel a distance of approximately 18,058.7 meters (or 18.06 kilometers) to reach her maximum speed.

Answer 2

Answer: 8981.72 m

Step-by-step explanation:

We can use the following equation to solve this problem:

`V^(2)=V_(o)^(2) + 2ad`

Where:

`V=965 (km)/(h) (1000 m)/(1 km) (1 h)/(3600 s)=268.05 m/s` is Kitty's final (maximum) speed

`V_(o)=0 m/s` is Kitty's initial speed (since it started from rest)

`a=4 m/s^(2)` is the acceleration

`d` is the traveled distance

Isolating
`d`:

`d=(V^(2)-V_(o)^(2))/(2a)`

`d=((268.05 m/s)^(2)-(0 m/s)^(2))/(2(268.05 m/s))`

Finally:

`d=8981.72 m`