Question

a car slows down uniformly from a speed of 21.0 m/s to rest in 6.00s. how far did it travel in that time?

Answer 1

The kinematic equation to find distance is:


`d = v(initial) . t + 0.5a . t^(2)`

where,
v = velocity
a = acceleration
t = time

Since we need to find 'a' (which is negative due to the car slowing down), use the equation a = Δv / t = -21 / 6 = -3.5m/s^2
so the equation looks like this:


`d = (v(initial) .t)+(0.5[v(delta)/t]).(t^(2))`

d = (21 x 6) + (0.5 x -3.5) 6^2 = 189meters

Answer 2

The traveled distance is 63.0 m

Step-by-step explanation:

We can write the equation for the acceleration as the variation of the velocity over time:

dv/dt = a

where

dv/dt = variation of velocity over time

a = constant acceleration

Then separating variables, we can solve the equation:

dv = a dt

Integrating between the initial velocity ( v0) and final velocity (v) and between t = 0 and t = t.

v - v0 = a · t

v = v0 + a · t

This is the equation of velocity at time t.

Now, we can write the velocity as the variation of the position over time (dx/dt). Then:

dx/dt = v0 + a · t

Separating varibles:

dx = v0 · dt + a · t · dt

Integrating between the initial position (x0) and final position (x) and between t = 0 and t

x -x0 = v0 · t + 1/2 · a · t²

x = x0 + v0 · t + 1/2 · a · t²

This is the equation of position at time t.

Now, we know that the velocity at time 6.00 s is 0. Then:

v = v0 + a · t

Solving for a

(v -v0)/t = a

(0- 21.0 m/s)/6.00s = -3.50 m/s² = a

Then, using this acceleration in the equation of the position:

x = x0 + v0 · t + 1/2 · a · t²

x = 0 m + 21.0 m/s · 6.00 s + 1/2 · (-3.50 m/s²) · (6.00 s)²

x = 63.0 m

The traveled distance is 63.0 m