Q10. Sitting in the passenger seat you observe that the drive of the car needs only 3 seconds to come from a speed of 35 mph on the entrance ramp to the interstate to a speed of 70 mph on the interstate. What is the car's acceleration? How much distance does the car cover during those 3 seconds?
(70-35)/3 = ___ mi/hr-s
s = vt + 1/2 at^2
s = 35/3600 mi/s * 3s + 1/2 (35/(3*3600)) mi/s^2 * (33 s)^2 = 7/160 mi
Well, it sounds like that driver has quite the need for speed! To find the car's acceleration, we can use the formula:
acceleration = (final velocity - initial velocity) / time
In this case, the final velocity is 70 mph, the initial velocity is 35 mph, and the time is 3 seconds. So let's do the math:
acceleration = (70 mph - 35 mph) / 3 sec
acceleration = 35 mph / 3 sec
acceleration ≈ 11.67 mph/sec
So the car's acceleration is approximately 11.67 mph/sec. That's like going from 0 to 60 in "I just remembered I left the oven on" seconds!
To calculate the distance the car covers during those 3 seconds, we can use another formula:
distance = (initial velocity * time) + (0.5 * acceleration * time^2)
Since we already know the initial velocity (35 mph), the time (3 seconds), and the acceleration (11.67 mph/sec), we can plug those values in:
distance = (35 mph * 3 sec) + (0.5 * 11.67 mph/sec * (3 sec)^2)
distance = 105 miles + 52.5 miles
distance = 157.5 miles
So during those 3 seconds, the car covers a distance of approximately 157.5 miles. That's like driving from "Are we there yet?" to "I thought we were already there!" in the blink of an eye!
To find the car's acceleration, we can use the formula:
acceleration (a) = (final velocity (v) - initial velocity (u)) / time (t)
In this case, the initial velocity (u) is 35 mph, the final velocity (v) is 70 mph, and the time (t) is 3 seconds.
Step 1: Convert the velocities from mph to m/s
1 mph is equal to 0.447 m/s.
Initial velocity (u) = 35 mph × 0.447 m/s ≈ 15.695 m/s
Final velocity (v) = 70 mph × 0.447 m/s ≈ 31.290 m/s
Step 2: Calculate the acceleration
acceleration (a) = (31.290 m/s - 15.695 m/s) / 3 s ≈ 5.198 m/s²
Therefore, the car's acceleration is approximately 5.198 m/s².
To find the distance covered by the car during those 3 seconds, we can use the formula:
distance (d) = initial velocity (u) × time (t) + 0.5 × acceleration (a) × time² (t²)
Step 1: Plug in the values
initial velocity (u) = 15.695 m/s
time (t) = 3 s
acceleration (a) = 5.198 m/s²
Step 2: Calculate the distance
distance (d) = 15.695 m/s × 3 s + 0.5 × 5.198 m/s² × (3 s)² ≈ 47.085 m
Therefore, the car covers approximately 47.085 meters during those 3 seconds.
To determine the car's acceleration, we can use the formula:
acceleration = (final velocity - initial velocity) / time
Given:
Initial velocity (u) = 35 mph
Final velocity (v) = 70 mph
Time taken (t) = 3 seconds
First, we need to convert the velocities from mph to m/s to use consistent units. Since 1 mph is approximately equal to 0.447 m/s, we have:
Initial velocity (u) = 35 mph * (0.447 m/s) = 15.615 m/s
Final velocity (v) = 70 mph * (0.447 m/s) = 31.290 m/s
Next, we can calculate the acceleration:
acceleration = (31.290 m/s - 15.615 m/s) / 3 seconds
acceleration = 15.675 m/s / 3 seconds
acceleration ≈ 5.225 m/s²
Therefore, the car's acceleration is approximately 5.225 m/s².
To find the distance covered by the car during those 3 seconds, we can use the formula:
distance = (initial velocity + final velocity) * (time / 2)
distance = (15.615 m/s + 31.290 m/s) * (3 seconds / 2)
distance = 46.905 m/s * 1.5 seconds
distance ≈ 70.358 meters
Hence, the car covers approximately 70.358 meters during the 3 seconds.