If a car accelerates at 4.0 m/s2, how long will it take to reach a speed of 80 km/hr, starting from rest?
You need to look at the units of your values and do the right cancellation to end up with what you need. So lets see if we can do that here: I would start by converting the 80km/hr into units of m/s so we can make the proper comparison to the acceleration units. What is 80 km in m? 80 km x 1000m/1km = 80,000 m What is 1 hr in s? 1 hr x 60 min/1 hr x 60 s/1 min = 3600 s So the speed we need to reach is 80,000 m / 3600 s = 22.22 m/s Now, what does it mean to accelerate? It means that we increase our speed by a certain number per unit time. We say it is the rate of change of velocity over time. We are talking about UNIFORM or constant acceleration which is a type of motion in which the velocity of an object changes by an equal amount in every equal time period. That's why our units are meters per second per second. So in your problem, we know we are increasing our speed by 4.0 m/s every second. So the question we ask ourselves is how many 4.0 values can go into the final speed of 22.22? This makes sense because this is the number were are trying to REACH. Using division we get, 22.22 m s^-1/4.0 m s^-2 = 5.6 seconds So if we start from stop.....and accelerate 4.0 m per second per second, it would take 5.6 seconds to reach a speed of 22.22 meters per second (80 km/hr).
A car is traveling 60 km/h when it begins to accelerate at a rate of 4.0 m/s^2. how long will it take the car?
a = (v-u)/t or t = (v-u)/a = (90-60)*(10/36) /4 = 2.0833 s <------ if we take 1 km/hr = 0.278m/s, t = 2.085s <------
A car is traveling 60 km/h when it begins to accelerate at a rate of 4.0 m/s^2. how long will it take the car?
a car is traveling 60 km/h when it begins to accelerate at a rate of 4.0 m/s^2. how long will it take the car to get 90 km/h ? hint 1Km/h = 0.278 m/s please help!!!
A car starting from rest accelerates at a constant rate of 3.0 m/s2 for 8.0 seconds. How far does the car travel during this time?
Apply newton's 2nd equation of motion which goes asS = u×t +1/2×a×(t^2)Where s is distanceu is initial velocitya is the accelerationt is the time of acceleration.In this problem the car starts from rest so u =0.Putting the other quantities in the equation we seeS = 1/2× 3.0× 8.0 × 8.0S = 96m.Special attention must be given to the units. As all the quantities must be first changed into matching units i.e cgs or s.i. units before applying the equation.
A car travelling at 30 m/s approaches 10m from an intersection. When the driver sees a pedestrian and slams on his brakes, and decelerates at a rate of 50 m/s, how long does it take the car to come to a stop?
If this is homework, please take a stab for yourself using only my first two equations. Assuming that “50 m/s deceleration” actually means “50 m/s^2 deceleration” —[math]v = at[/math][math]d = \frac{1}{2} at^2[/math]And thus[math]d = \frac{v^2}{2a}[/math][math]d = \frac{(30)^2}{(2 * 50)}[/math][math]d = \frac{900}{100}[/math][math]d = [/math]9 metersWhich works out rather poetically… the car stops 9 meters into its 10-meter danger zone, a scant one meter away from the presumably-terrified pedestrian.
If a car accelerates from rest to 25 m/s, how long does it take to this?
that very much depends on the acceleration, if the acceleration is constant, then the time will be proportional to the speed (25m/s in your case) and reciprocal as the acceleration, however, if the car accelerates as fast as possible the acceleration will never be constant and the only way to know then is to look at the velocity-time profiles.
A car has a velocity of 10m/s. It now accelerate at 4m/s^2 for a half a minute. What is the distance travelled in this time and the final speed of the car?
For final speed, use:[math]v = u + at[/math][math]v = 10m/s + 4m/s^2*30s[/math][math]v = 10m/s + 120m/s[/math][math]v = 130m/s[/math]For distance, use:[math]s = ut + 1/2at^2[/math][math]s = 10m/s*30s + 1/2*4m/s^2*(30s)^2[/math][math]s = 300m + 2m/s^2*900s^2[/math][math]s = 300m + 1800m[/math][math]s = 2100m[/math]Thus the final sped of the car will be 130m/s and the distance covered is 2100m
A car starts from rest and accelerates at 6 m/s. How far does it travel in 3.00 s?
27 m.You mean 6 m/s^2 for acceleration.After 3 s, the velocity went from 0 to 18 m/s. The average velocity during this 3 s is 18/2 = 9 m/s. The distance traveled at an average velocity of 9 m/s for 3 s is 9•3 = 27 m.