A driver in a car traveling at a speed of 21.8 m/s sees a car 101 m away on the road. How long will it take for the car to uniformly decelerate to a complete stop in exactly 99 m?

Question

Benjamin Asiedu · Accepted Answer

#9.1s#

Since we know that velocity is uniformly decelerating, we can take the average velocity:

#(V_i + V_f)/2#

Letting #V_f# equal 0 m/s, we get #1/2*21.8m#/#s# or #10.9m#/#s#.

Now we know #V = d/t# and rearranging this gives #t = d/V#.

Substitute our velocity of 10.9m/s for V and 99m for d: #t=(99m)/(10.9m*s^-1)# (Sorry for the #s^-1# but otherwise I couldn't write m/s)

Finally we get #t=9.1s# (with significant figure rounding).

Hope this helps!

Cheers

Parker Abbott · Answer

undefined To find the time it takes for the car to decelerate to a complete stop, we can use the equation of motion for uniformly decelerated motion:

$$ v^2 = u^2 + 2a s $$

Where:
- $ v $ is the final velocity (0 m/s, as the car comes to a stop),
- $ u $ is the initial velocity (21.8 m/s),
- $ a $ is the acceleration (which is negative since the car is decelerating),
- $ s $ is the displacement (99 m, the distance over which the car decelerates).

Rearranging the equation to solve for acceleration ($ a $), we get:

$$ a = \frac{{v^2 - u^2}}{{2s}} $$

Substituting the given values, we have:

$$ a = \frac{{0^2 - (21.8)^2}}{{2 \times 99}} $$

$$ a = \frac{{-475.24}}{{198}} $$

$$ a ≈ -2.4 \, \text{m/s}^2 $$

Now, we can use the equation of motion for uniformly accelerated motion to find the time ($ t $) it takes for the car to decelerate to a stop:

$$ v = u + at $$

Substituting the known values:

$$ 0 = 21.8 - 2.4t $$

Solving for $ t $:

$$ 2.4t = 21.8 $$

$$ t = \frac{{21.8}}{{2.4}} $$

$$ t ≈ 9.083 \, \text{s} $$

Therefore, it will take approximately 9.083 seconds for the car to uniformly decelerate to a complete stop over a distance of 99 meters.