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As the acceleration is uniform, the train has an average speed that is half the difference between the start and final velocities, which in this case is half the final velocity.

1 hr = 60 min

1 km/h = 1 km ÷ 1 hr

= km ÷ 60 min

= 1/60 km/min

Distance = velocity × time

= (½ × 72 × 1/60 km/min) × (5 min)

= 36/60 × 5 km

= 3 km

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3 kilometres.

Q: A train starting from rest attains a velocity of 72Km hr in 5 minutes. Assuming that the acceleration is uniform what is the distance travel by train while attained the velocity?

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There are 3 formula 1. Final velocity = starting velocity + (acceleration)(time) 2. Final velocity^2 = starting velocity^2 + 2(acceleration)(distance) 3. Distance = (starting velocity)(time) + 1/2(acceleration)(time^2) Use whichever you can use.

If starting from rest, Distance = 1/2 (acceleration) x (time)2 . Otherwise, Distance = 1/2 (initial speed + final speed) x (time)

You can use the formula for distance covered:distance = (initial velocity) x (time) + (1/2) (acceleration) (time squared) Solve for time. This assumes constant acceleration, by the way. If you assume that the initial velocity is zero, then you can omit the first term on the right. This makes the equation especially easy to solve.

Acceleration is the rate of change in velocity, that is the finishing velocity minus the starting velocity divided by the time taken for that change. Velocity is the rate of change in distance, that is the finishing distance minus the starting distance divided by the time taken for that change. Distance may be measured in metres and time in seconds. In that case velocity would be measured in metres per second. Acceleration is then measured in (metres per second) per second or metres per second2.

the final velocity assuming that the mass is falling and that air resistance can be ignored but it is acceleration not mass that is important (can be gravity) final velocity is = ( (starting velocity)2 x 2 x acceleration x height )0.5

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There are 3 formula 1. Final velocity = starting velocity + (acceleration)(time) 2. Final velocity^2 = starting velocity^2 + 2(acceleration)(distance) 3. Distance = (starting velocity)(time) + 1/2(acceleration)(time^2) Use whichever you can use.

If starting from rest, Distance = 1/2 (acceleration) x (time)2 . Otherwise, Distance = 1/2 (initial speed + final speed) x (time)

You can use the formula for distance covered:distance = (initial velocity) x (time) + (1/2) (acceleration) (time squared) Solve for time. This assumes constant acceleration, by the way. If you assume that the initial velocity is zero, then you can omit the first term on the right. This makes the equation especially easy to solve.

Look at the metric ruler pictured. Assuming that 0 is visible on the measure and is also the starting point, what is the distance to point D?

Acceleration is the rate of change in velocity, that is the finishing velocity minus the starting velocity divided by the time taken for that change. Velocity is the rate of change in distance, that is the finishing distance minus the starting distance divided by the time taken for that change. Distance may be measured in metres and time in seconds. In that case velocity would be measured in metres per second. Acceleration is then measured in (metres per second) per second or metres per second2.

parabolic graphs where f(t)=xi + vt + 1/2at^2 f(t) = distance travelled, xi = starting distance, v = starting velocity, t = time elapsed, and a = acceleration.

Formula for distance traveled with constant acceleration is: S = v0t + at2 / 2. Knowing the distance, time and that initial velocity is zero: S = at2 / 2, a = 2S / t2, or numerically: a = 804 / 36 = 22.33 m/s2 Expressing that in terms of g(=9.81 m/s2): a = 2.28 g

5.5 cm

the final velocity assuming that the mass is falling and that air resistance can be ignored but it is acceleration not mass that is important (can be gravity) final velocity is = ( (starting velocity)2 x 2 x acceleration x height )0.5

One example... X = 1/2 A t2 + V0 t + X0 Where X is distance, A is acceleration, t is time, V0 is initial velocity, and X0 is initial distance. This allows you to calculate where you would be given a starting position, velocity, and acceleration, after a specified time, such as in an automobile.

Distance from starting point Instantaneous velocity Average velocity Acceleration or deceleration Rate of change of acceleration and higher rates of change. Some of these can only be determined if the diagram is smooth or for smooth parts of the diagram.

Will you settle for average speed ?The general method for solving any problem is to use what you do know in orderto find the answer. Generally, you never use something you don't know, right ?Well, you told us what you don't have, but neglected to mention what you do have.If you know starting and ending speed, then average speed = 1/2 (starting speed + ending speed).If you know distance and acceleration, then time = sqrt( 2 x distance / acceleration).Then you have time and distance, and you can find average speed.