Instruction

1

Use the formula F=ma, where F is the force (measured in Newtons), m is the mass of the vehicle, a is the acceleration. To find mass, use the rule to find unknown multiplier: "to find an unknown multiplier, we need work divided by the known multiplier." Work: m=F/a.

2

Now replace the known value of acceleration – speed (V). Use the formula a=V/t, where t is the time in which the rise of the machine. If the time is given in seconds, speed in meters per minute, that may equalize the values. Put time in minutes, speed in meters per second.

3

In the original formula m=F/a substitute the resulting value of acceleration. Work: m=F/V/t. Use the rule of division by a fraction: "When you divide by a fraction, its denominator goes up, the numerator down." Hence: m=Ft/V.

4

Now, to find mass, substitute into the formula m=Ft/V known values. For example: F=50 N (Newtons), t=10 s (seconds), V=1 m/s (meters per second). Work: m=50 N x 10 s / 1 m/s, m=500 kilograms.

# Advice 2: How to find the time, knowing the speed

Tasks on kinematics, which is necessary to calculate

**speed**,**time**or path of a uniformly and rectilinearly moving bodies, meet in the school course of algebra and physics. For their solution find it in the condition values, which can be interconnected to equalize. If the condition is required to determine**the time**at a known speed, use the following instructions.You will need

- - handle;
- paper for records.

Instruction

1

The simplest case is the motion of a body with a given uniform

**velocity**u. Know the distance that the body has passed. Find**time**in transit: t = S/v, h, where S is distance, v is the average**velocity**of the body.2

The second example is for a counter motion. From point A to point b the car is moving with

**speed**u 50 km/h. to Meet him from point B at the same time I left the moped with**the speed**u 30 km/h. The distance between the points A and b 100 km is Required to find**the time**through which they will meet.3

Designate a meeting point with the letter K. Let the distance of AK, who drove the car be x km. Then the path of the rider will amount to 100's of km. From the problem statement it follows that

**the time**in the way of car and motorbike alike. Write down the equation: x/v = (S-x)/v’, where v, v’ – vehicle speed, and a moped. Substituting the data solve the equation: x = 62,5 km. Now find**the time**: t = for 62.5/50 = 1.25 hours or 1 hour and 15 minutes.4

The third example – given the same conditions, but left 20 minutes late the bike. To determine how much time will the car before meeting with a moped.

5

Write down the equation, similar to the previous one. But in this case

**time**of a moped in the path will be 20 minutes more than the car. For the adjustment of the parts, subtract one-third of the hours from the right side of the expression: x/v = (S-x)/v’-1/3. Find the x – 56,25. Calculate**time**: t = 56,25/50 = 1,125 hours, or 1 hour 7 minutes 30 seconds.6

The fourth example is a challenge for a body moving in one direction. Car and moped with the same speed move from point A. it is Known that the car left half an hour later. After a

**time**he will catch up with the moped?7

In this case, the same will be the distance traveled of the vehicle. Let

**time**in the path of the car be x hours then**time**in the path of the moped will be x+0,5 hours. You got the equation: vx = v’(x+0,5). Solve the equation, substituting the value of speed, and find x – 0.75 hour or 45 minutes.8

The fifth example is a car and a moped with the same speed moving in the same direction, but the moped went from point b, located at a distance of 10 km from point A, half an hour earlier. To calculate, through what

**time**after the start of the car will overtake the moped.9

The distance that the car drove 10 km more. Add this difference to the path of the motorcycle and align the parts of the expression: vx = v’(x+0,5)-10. Substituting the values of speed and solved it, you will get the answer: t = 1.25 hours or 1 hour and 15 minutes.