There are a few ways to go about solving this one. Here's one that takes a few steps, but they're easy to follow. What is the average speed of the car during its rapid deceleration? Assuming constant deceleration, it's (80 - 0)/2 = 40 km/s = 11.1 m/s. (I'm playing fast and loose with the signs at this point.) Divide the stopping distance (50 m) by that speed and you get 4.5 seconds. Acceleration, a, is delta V / delta t = -22.2 / 4.5 = -4.9 m/s2. Since the acceleration of gravity is -9.8 m/s2, the deceleration of the car is about half a g. Note that deceleration is the same as negative acceleration.
Chat with our AI personalities
Acceleration = 7.09 ms-2 G's = 7.09/9.8 = 0.72
Acceleration is a change in velocity per unit of time. Velocity is distance (d) per unit of time (t). That makes acceleration distance per unit of time squared, or something like this:We have distance/time2, or d/t2Distance is commonly measured in meters, and time in seconds. This makes acceleration appear in meters per second per second, or meters per second squared, or m/sec2.m/s2meters per second squared
9.8 meters per second squared is the acceleration of gravity.
That's unusual. I guess your teacher is trying to make you think a bit. It's a good mental exercise, though. You may recall that the units of acceleration are meters per second squared. That gives you a clue right there. And if you knew Calculus, you'd know that acceleration is the second derivative of distance, s, with respect to time, t: d2s/dt2. So, by now you're probably getting the feeling that the slope of a distance-time squared graph has something to do with acceleration. And you'd be right. Just as the slope of a velocity-time graph is acceleration, the slope of a distance-t2 graph is acceleration. Well, not quite. It's actually ONE HALF the acceleration.
The sprinter's acceleration is 2 meters per second squared.