According to recent test data, an automobile travels 0.250 mi in 19.9 s, starting from rest. The same car, when braking from 60.0 mi/h on dry pavement, stops in 146 ft. Assume constant acceleration in each part of the motion, but not necessarily the same acceleration when slowing down as when speeding up.

Question

According to recent test data, an automobile travels 0.250 mi in 19.9 s, starting from rest. The same car, when braking from 60.0 mi/h on dry pavement, stops in 146 ft. Assume constant acceleration in each part of the motion, but not necessarily the same acceleration when slowing down as when speeding up.

If its acceleration is constant, how fast (in mi/h) should this car be traveling after 0.250 mi of acceleration? The actual measured speed is 70.0 mi/h what does this tell you about the motion?

Answer 1

Ab = V²/2x = -88²/(2*146) = -26.52 ft/sec²

Aa = 2x/t² = 2(5280/4)/19.9² = 6.666 ft/sec²

V = √[2ax] = √[2*6.666*5280/4] = 132.658 ft/sec = 90.4 mph

t = √[2x/a] = √[2*5280/(4*26.52)] = 9.98 sec

Answer 2

To find the final speed of the car after 0.250 mi of acceleration, we can use the kinematic equation:

v^2 = u^2 + 2as

where v is the final speed, u is the initial speed (which is 0 since the car starts from rest), a is the acceleration, and s is the distance traveled.

Let's find the acceleration first using the given data. The car travels 0.250 mi in 19.9 s, so the distance is 0.250 mi and the time is 19.9 s.

s = 0.250 mi
t = 19.9 s

Using the equation:

s = ut + 0.5at^2

0.250 = 0 + 0.5a(19.9)^2
0.250 = 0.5a * 396.01
a = 0.250 / (0.5 * 396.01)
a ≈ 0.00126 mi/s^2

Now, let's calculate the final speed using the same kinematic equation:

v^2 = u^2 + 2as

v^2 = 0^2 + 2 * 0.00126 * 0.250
v^2 = 0 + 0.00063
v ≈ 0.02508 mi/s

To convert this speed from miles per second to miles per hour:

v (mi/h) = v (mi/s) * 3600 (s/h)
v (mi/h) ≈ 0.02508 * 3600
v (mi/h) ≈ 90.288 mi/h

So, the car should be traveling at approximately 90.288 mi/h after accelerating for 0.250 mi.

Given that the actual measured speed is 70 mi/h, we can conclude that the car has exceeded the expected speed. This suggests that there might be other factors involved in the car's motion, such as external forces or a different acceleration profile.

Answer 3

To find the speed of the car after 0.250 mi of acceleration, we can use the equation of motion:

v^2 = u^2 + 2as

Where:
v = final velocity
u = initial velocity (0 in this case as the car starts from rest)
a = acceleration
s = displacement (0.250 mi in this case)

Rearranging the equation, we get:

v = √(2as)

Substituting the known values, we get:

v = √(2 * a * 0.250)

Now we need to find the value of acceleration (a). We can use the given test data:

Distance, s = 0.250 mi
Time, t = 19.9 s

Using the equation of motion:

s = ut + (1/2)at^2

Since initial velocity (u) is 0, the equation simplifies to:

s = (1/2)at^2

Rearranging the equation, we get:

a = (2s) / t^2

Substituting the known values, we get:

a = (2 * 0.250) / (19.9^2)

Now we have the value of acceleration (a), which we can substitute into the equation for velocity:

v = √(2 * a * 0.250)

Calculating this will give you the speed of the car after 0.250 mi of acceleration.

Now, let's discuss the second part of the question. The actual measured speed is given as 70.0 mi/h. Since the calculated speed is different (according to the given test data), it tells us that the car does not maintain constant acceleration throughout its motion. The acceleration while slowing down is different from the acceleration while speeding up. This suggests that the car may have a different acceleration profile during deceleration (braking) compared to acceleration.