A VW Beetle goes from 0 to 59.0 mi/h with an acceleration of +2.35 m/s2. (a) How much time does it take for the Beetle to reach this speed? (b) A top-fuel dragster can go from 0 to 59.0 mi/h in 0.700 s. Find the acceleration (in m/s2) of the dragster.
To find the time it takes for the Beetle to reach a speed of 59.0 mi/h, we can use the equation of motion:
v = u + at
where:
- v is the final velocity (59.0 mi/h)
- u is the initial velocity (0 mi/h or 0 m/s)
- a is the acceleration (2.35 m/s^2)
- t is the time taken
First, let's convert the final velocity from mi/h to m/s:
59.0 mi/h * (1.609 km/mi) * (1000 m/km) * (1 h/3600 s) = 26.36 m/s
Now we can substitute the values into the equation and solve for t:
26.36 m/s = 0 m/s + 2.35 m/s^2 * t
Simplifying the equation:
26.36 m/s = 2.35 m/s^2 * t
Dividing both sides by 2.35 m/s^2:
t = 26.36 m/s / 2.35 m/s^2
t ≈ 11.2 seconds
Therefore, it takes approximately 11.2 seconds for the Beetle to reach a speed of 59.0 mi/h.
To find the acceleration of the top-fuel dragster, we can rearrange the equation of motion:
v = u + at
to solve for acceleration (a):
a = (v - u) / t
where:
- v is the final velocity (59.0 mi/h)
- u is the initial velocity (0 mi/h or 0 m/s)
- t is the time taken (0.700 s)
Let's convert the final velocity from mi/h to m/s:
59.0 mi/h * (1.609 km/mi) * (1000 m/km) * (1 h/3600 s) = 26.36 m/s
Substituting the values into the equation:
a = (26.36 m/s - 0 m/s) / 0.700 s
Simplifying the equation:
a = 26.36 m/s / 0.700 s
a ≈ 37.66 m/s^2
Therefore, the acceleration of the top-fuel dragster is approximately 37.66 m/s^2.