A car is traveling 20 m/s when the driver sees a child standing on the road. He takes 0.8 s to react, then steps on the brakes and slows at 7.0 m/s2. How far does the car go before it stops?
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To determine the distance the car goes before stopping, we need to calculate two separate distances: the distance traveled during the reaction time and the distance traveled while slowing down.
1. First, let's calculate the distance traveled during the reaction time. The formula to calculate distance is:
distance = initial velocity * time + 0.5 * acceleration * time^2
Given:
initial velocity (v0) = 20 m/s (the car's speed)
time (t) = 0.8 s (reaction time)
acceleration (a) = 0 m/s^2 (since the car is not accelerating during the reaction time)
Plugging in these values, we get:
distance_reaction = v0 * t + 0.5 * a * t^2
= 20 * 0.8 + 0.5 * 0 * (0.8)^2
= 16 + 0
= 16 meters
Therefore, the distance traveled during the reaction time is 16 meters.
2. Next, let's calculate the distance traveled while slowing down. To do this, we will use the equation:
velocity_f^2 = velocity_i^2 + 2 * acceleration * distance
We know that the final velocity (velocity_f) will be 0 m/s since the car comes to a stop. Therefore, we can rearrange the equation to solve for distance:
distance_slowing = (velocity_f^2 - velocity_i^2) / (2 * acceleration)
Given:
initial velocity (v0) = 20 m/s (the car's speed)
final velocity (vf) = 0 m/s (stopped)
acceleration (a) = -7.0 m/s^2 (negative sign indicates deceleration)
Plugging in these values, we get:
distance_slowing = (0^2 - 20^2) / (2 * -7.0)
= -400 / (-14)
= 28.57 meters (rounded to two decimal places)
Therefore, the distance traveled while slowing down is approximately 28.57 meters.
3. Finally, we can find the total distance traveled by adding the distances traveled during the reaction time and while slowing down:
total distance = distance_reaction + distance_slowing
= 16 + 28.57
= 44.57 meters (rounded to two decimal places)
Therefore, the car goes approximately 44.57 meters before it stops when the driver sees the child on the road.
reaction time = 0.8 seconds at 20 m/s
distance = 0.8 s * 20 m/s = 16 m.
From v0²-v² = 2aS
where v0 and v are initial and final velocities,
a, acceleration
S, distance travelled
braking distance
= (0-20²)/(2*(-7 m/s²))
= 40.8 m.
Total distance = ?