a rocket propeled sled moved along a track at 1020km/h He and the sled were brought to a stop in 1.4s In g units what acceleration did he e xperiance while stopping?
I know 1 g = 9.8m/s^2
I used a=(v-v0)/t and got 74.344 g
Is this correct?
Acceleration, or in this case deceleration, is (v_f - v_0)/time
v_0=1020km/h, v_f=0, t=1.4sec
Are you sure it's 1020km/hr? That seems a bit high.
1020km/hr=1,020,000m/3600sec=283.3m/sec
So a=-283.3m/sec/1.4sec=approx 202m/sec^2
202/9.8=approx 20.7g
Check my arithmetic. I'm not sure how you got your answer.
To calculate the acceleration, or in this case deceleration, of the rocket-propelled sled, you correctly used the formula a = (v_f - v_0) / t, where v_f is the final velocity (0 in this case), v_0 is the initial velocity (1020 km/h), and t is the time taken to come to a stop (1.4 seconds).
However, it seems there was a mistake in your calculation. Let's break it down step by step to find the correct answer.
1. Convert the initial velocity from km/h to m/s:
v_0 = 1020 km/h × (1000 m/1 km) / (3600 s/1 h) = 283.3 m/s (rounded to one decimal place)
2. Calculate the acceleration using the formula:
a = (v_f - v_0) / t
= (0 - 283.3 m/s) / 1.4 s
= -202.4 m/s² (rounded to one decimal place)
3. Convert the acceleration from m/s² to g-units:
g = 1 / 9.8 m/s² ≈ 0.102 g
acceleration in g-units = -202.4 m/s² × 0.102 g = -20.7 g (rounded to one decimal place)
Therefore, the correct answer is approximately -20.7 g, indicating that the person experienced a deceleration of about 20.7 times the acceleration due to gravity while stopping.
Note: The negative sign indicates that the acceleration is in the opposite direction of the initial velocity, which makes sense for deceleration.