An experimental rocket car starting from rest reaches a speed of 600 km/h after a straight 370 m run on a level salt flat. Assuming that acceleration is constant, answer the following questions.
a)What was the time of the run? (in seconds)
b)What is the magnitude of acceleration? (in m/s^2)
first convert 600km/h to m/s then use the equation
final velocity suqared = initial velocity squared plus 2 times aceleration times distance. remember initial is 0m/s. Then once you find acceleration use the equation final velocity equals initial velocity plus acceleration times time to solve for the time.
To solve these questions, we need to use the equations of motion for constant acceleration.
a) What was the time of the run? (in seconds)
We can use the equation:
v = u + at
Where:
v = final velocity (600 km/h)
u = initial velocity (0 m/s, since the car starts from rest)
a = acceleration (to be determined)
t = time
First, we need to convert the final velocity from km/h to m/s:
600 km/h × (1000 m/1 km) × (1 h/3600 s) = 166.67 m/s
Now we can rewrite the equation as:
166.67 = 0 + a × t
Since the initial velocity is 0, the equation simplifies to:
166.67 = a × t
To solve for t, we need to find the acceleration a. We can use another equation:
v^2 = u^2 + 2as
Where:
v = final velocity (166.67 m/s)
u = initial velocity (0 m/s)
a = acceleration (to be determined)
s = displacement (370 m)
Rearranging the equation to solve for a, we get:
a = (v^2 - u^2) / (2s)
a = (166.67^2 - 0^2) / (2 × 370)
a = 13888.9 m/s^2
Now we can substitute the acceleration back into the first equation to solve for t:
166.67 = 13888.9 × t
Solving for t:
t = 166.67 / 13888.9
t ≈ 0.012 seconds
Therefore, the time of the run is approximately 0.012 seconds.
b) What is the magnitude of acceleration? (in m/s^2)
We have already calculated the acceleration in part a) as 13888.9 m/s^2. So, the magnitude of acceleration is 13888.9 m/s^2.