Question

67. Race Car A race car is slowed with a constant accel-

eration of 11 m/s2 opposite the direction of motion.
a. If the car is going 55 m/s, how many meters will it
travel before it stops?
b. How many meters will it take to stop a car going
twice as fast?

Answer 1

a. 137.5 m

b. 550 m

Step-by-step explanation:

Accelerated Motion

If an object is changing its velocity at a constant rate, it has a uniformly accelerated motion. When the object is moving in one fixed axis, then the sign of the acceleration is negative if the object is braking, and positive if the object is increasing its speed.

The initial speed vo, final speed vt, acceleration a, and distance traveled x are related by the formula

`v_f^2=v_o^2+2ax`

a. The Race Car A has an initial speed of 55 m/s and it's said to stop. We must find at what distance it goes to vf=0. This means that the above formula becomes

`0=v_o^2+2ax`

Solving for x

`\displaystyle x=-(v_o^2)/(2a)`

The acceleration is
`-11\ m/s^2`, negative because it's against the movement. Thus

`\displaystyle x=-(55^2)/(2* (-11))`

`x=137.5\ m`

b. If the car is going twice as fast (v0=110 m/s), then

`\displaystyle x=-(110^2)/(2* (-11))`

`x=550\ m`