Question
A \( 1200-\mathrm{kg} \) automobile moving at \( 25 \mathrm{~m} / \mathrm{s} \) has the brakes applied with a deceleration of \( 8.0 \mathrm{~m} / \mathrm{s}^{2} \). How far does the car travel before it stops? 1) 39 m 2) 47 m 3) 55 m 4) 63 m 5) 72 m
Ask by Harper Welch. in the United States
Feb 25,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
The car travels approximately 39 meters before it stops.
Solution
We can solve the problem using the kinematic equation that relates velocity, acceleration, and displacement:
v² = v₀² + 2a x
Here,
v = final velocity = 0 m/s,
v₀ = initial velocity = 25 m/s,
a = acceleration = -8.0 m/s² (the negative sign indicates deceleration),
x = distance traveled.
Plugging in the numbers:
0 = (25)² + 2(–8.0) x
0 = 625 – 16x
Solve for x:
16x = 625
x = 625 / 16
x ≈ 39.06 m
Thus, the car travels approximately 39 m before it stops.
The correct answer is 1) 39 m.
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Simplify this solution Beyond the Answer
To find the distance the car travels before it stops, we can use the kinematic equation: \[ v^2 = u^2 + 2as \] where: - \( v \) = final velocity (0 m/s since the car stops), - \( u \) = initial velocity (25 m/s), - \( a \) = acceleration (which is negative in this case due to deceleration, so \( -8 \, \mathrm{m/s^2} \)), and - \( s \) = distance traveled. Plugging in the values: \[ 0 = (25)^2 + 2(-8)s \] This simplifies to: \[ 0 = 625 - 16s \] Rearranging the equation gives: \[ 16s = 625 \\ s = \frac{625}{16} \\ s = 39.0625 \, \mathrm{m} \] So, rounding we find that the car travels approximately 39 m before it stops. This corresponds to option 1) 39 m.
