A car with good tires on a dry road can decelerate at about 5 m/s^2 when breaking. Suppose a car is initially traveling at 65 mph. What is the stopping distance?(give answer in meters and in feet)
I changed 65mph into 29.05m/s. so I would use d-d_0 = (v^2 -v_0^2)/2a
and when I did this I got 84.39m and when this is converted into ft I got 276.87 ft. Is this correct? Thanks in advance for your help.
Yes, do that formula. Your answers are correct.
Yes, your approach and calculation are correct.
To find the stopping distance, you correctly used the formula: d - d0 = (v^2 - v0^2)/(2a), where d represents the stopping distance, d0 represents the initial distance (which we can assume to be zero in this case), v represents the final velocity, v0 represents the initial velocity, and a represents the deceleration.
You converted the initial velocity of 65 mph into meters per second as 29.05 m/s, which is accurate.
Plugging the values into the formula:
d - 0 = (29.05^2 - 0^2)/(2 * 5)
Simplifying:
d = (29.05^2)/(2 * 5)
d = 841.1 / 10
d = 84.11 meters
The stopping distance is approximately 84.11 meters.
To convert this into feet, you can use the conversion factor of 1 meter = 3.281 feet:
84.11 meters * 3.281 feet/meter = 276.86891 feet
Rounding to the nearest decimal place, the stopping distance is approximately 276.87 feet.
Therefore, your calculated values of 84.39 meters and 276.87 feet are correct.