How would you calculate the acceleration of a car (in km/h*s) that can go from rest to 100 km/h in 10 s?
The formula for Acceleration is Acc= (final velocity - initial velocity) / time Your final velocity (the speed you reach) is 100 km/h. Your initial speed (speed you start with) is 0, rest. The time you have is 10 seconds. So plug that into the formula. Acc= (FV - IV) / time Acc= (100 - 0) / 10 Solve that ^^ and there's your answer (:
How to determine the acceleration of a car during the braking period?
s = ut + 1/2 at^2; v = u+at; therfore s (distance) = (u +v )t/2 (initial velocity + final velocity)* time / 2 t = 2s / (u+v) = 2 * 50 / (20 +0) = 5 seconds deceleration = change in velocity/ change in time = 20/5 = 4m/s Use graph paper, time as x axis (in seconds), velocity as y axis (in metres), parallell straight line (at the 20m/s mark) until point of braking, then draw a line with a gradient of 4m/s, i.e. go down 4 units for every one unit across 60mph = 26.82 m/s. distance = (26.82 + 0)*4 / 2 = 53.64 metres Hope this helps!
Determine the acceleration a of the car?
A car is climbing the hill of slope theta1 at a constant speed v. if the slope decreases abruptly to theta2 at point A, determine the acceleration a of the car just after passing point A if the driver does not change the throttle setting or shift into a different gear.
Determine the acceleration of a car of mass 900kg,when a net force of 2700N act on it?
you can to the right place! f=ma thus a=f/m =2700/900 = 3 m/s^2
A car starting from rest accelerates to 50.0 m/s in 5.00 seconds. How do you find the acceleration and the distance covered by the car?
The problem has an ill formed conditions. There nowere said about the a(t) dependence. It coud be a good case when it sounds like “A car starting from rest accelerates to 50.0 m/s uniformely”, f.e.So the minim distance is zero meters. That accords to the variant when the car stood at rest all the 5 seconds and accelerates in a last moment so as its speed reaches the 50 m/s value. Last a moment can be a very small but a finite span to prevent infinity acceleration to be needed but provides the almost zero distance covered.And the maximum distance is 250 m/s by the very similar reason. This is, car reaches 50 m/s during the very small time from begining a moment and is moving all the 5 seconds with the constant speed.
A jogger accelerates from rest to 2.9 m/s in 2.6 s. A car accelerates from 37.0 to 44.0 m/s also in 2.6 s.?
(a) Find the acceleration (magnitude only) of the jogger. (b) Determine the acceleration (magnitude only) of the car. (c) Does the car travel farther than the jogger during the 2.6 s? If so, how much farther?
If a car accelerates from 0 to 100km/hr in 1.8 seconds, then how much distance is covered by that car in 1.8 seconds?
S=ut+1/2at^2S= distanceu = initial velocity = 0t = time = 1.8 sa = acceleration, which we assume to be constant.so S = 0 + 1/2(1.8^2) a= 1.62 (seconds squared) times aSo the distance is 1.62 times the acceleration.What’s the acceleration? The change in velocity over time.So (27.78 m/s) / 1.8s = 15.4 m/s^21.62 s^2 x 15.4 m/s^2 = 24.95 m - or about 25 m
What is the acceleration of a car that is slowing down? Suppose the car is slowing down at 30 km/hour, and finally stops after three seconds, what is the acceleration?
The acceleration of an object is the rate at which the speed changes. The acceleration of an object that is slowing down is negative, because the speed is decreasing.If the car starts out at 30km/hr and then takes 3 seconds to slow to 0, then it's average acceleration is -10km/hr*s (that's kilometers per hour per second- the rate of change of speed).That's average acceleration, but the acceleration of the car is not going to be constant. The brakes take time to reach full braking force, and the effectiveness of brakes decreases with the speed of the car.If you want the acceleration at one specific moment, you'll need to actually measure the speed of the car as it brakes, as the curve will vary from vehicle to vehicle.My guess is that you want the average acceleration.That's -10 km/(hr*s) or -1/360 km/s^2.
How do I calculate the acceleration of a car from the rest that travels at 24m/s due east in 30s?
Well, since you noted the direction of travel is “due east”, that clearly means the question is asking about a frame of reference that included planetary rotation into the mix. At the equator, that’s about 460m/s in the eastward direction. Clearly, at the north pole, that’s zero—you’re effectively stationary. So, let’s assume you’re somewhere between the equator and the pole, so we’ll take the average of the two, and say the earth is contributing 230m/sec of eastward velocity. So, at time t=0s, your velocity is 230m/sec; at time t=30s, your velocity is 254m/sec. Thus, you plug those into the formula, and you get (254–230)/(30–0), or 0.8m/sec^2.Now, if the question was to consider this from a larger frame of reference, we’d also have to take the rotation of the earth around the sun into consideration, which is about 107,000km/hr, or about 29,722m/sec. The problem is that we don’t know if we need to *add* that to the rotational velocity of the earth, and motion of the car, or *subtract* it; that all depends on whether the side of the earth that the car is on is facing the sun, or away from the sun. If we assume that sane people do math experiments on their cars only when the sun is shining, then we need to add the velocity in as well; so we get 29,722+230+24, or (29,976–29,952)(30–0). However, that only works if you do your vehicular calculations in the daytime. If, on the other hand, you’re a dark, brooding vigilante who only comes out after darkness falls to drive around, then we need to adjust our calculations to account for the fact that you’re now going retrograde with respect to the sun. Thus, the calculation would become 29,722–230–24, or (29468–29492)/(30–0), or -0.8m/sec^2.So, fundamentally, this problem comes down to one simple question:Are you Batman?
What is the acceleration of the car when it is in motion?
Why can't you just use F=ma? Newton's second law of motion says that the force acting on an object is proportional to the objects acceleration, given that the mass is constant. So F=ma. So a=F/m. Well F is 49.2 N and m is 19.3 kg. So a is 49.2/19.3 = 2.54922279793, which rounds to 2.55 ms^-2.