# Video: Finding the Distance Covered by a Car Moving with Uniform Acceleration

A body was moving in a straight line accelerating uniformly. If it covered 55 meters in the first 4 seconds and 57 meters in the next 4 seconds, determine the total distance covered in the first 10 seconds of its motion.

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### Video Transcript

A body was moving in a straight line accelerating uniformly. If it covered 55 meters in the first four seconds and 57 meters in the next four seconds, determine the total distance covered in the first 10 seconds of its motion.

In order to solve this problem, we will use some of the equations of motion or SUVAT equations. We will use 𝑠 equals 𝑢𝑡 plus half 𝑎𝑡 squared, 𝑠 equals 𝑣𝑡 minus half 𝑎𝑡 squared, and 𝑣 equals 𝑢 plus 𝑎𝑡, where 𝑠 is the displacement, 𝑢 is the initial velocity, 𝑣 is the final velocity, 𝑎 is the acceleration, and 𝑡 is the time. We can see from the diagram that in the first four seconds the body covered 55 meters. In the next four seconds, it covered 57 meters. We need to calculate the total distance covered in the first 10 seconds.

If we let the velocity of the body at 𝑡 equals four be 𝑣 meters per second, then for the first part of the journey we have a displacement of 55, we have a final velocity of 𝑣 and acceleration of 𝑎 and a time of four seconds. We can substitute these values into the equation 𝑠 equals 𝑣𝑡 minus a half 𝑎𝑡 squared. This gives us 55 is equal to 𝑣 times four minus a half multiplied by 𝑎 multiplied by four squared.

Simplifying this equation gives us 55 is equal to four 𝑣 minus eight 𝑎. This means that during the time period 𝑡 equals zero to 𝑡 equals four, the motion of the body satisfies the equation four 𝑣 minus eight 𝑎 equals 55.

If we now consider the time period between 𝑡 equals four and 𝑡 equals eight, we have a displacement of 57 meters. The initial velocity is equal to 𝑣 as this is the velocity at time equals four. The acceleration is still equal to 𝑎. And our time is four seconds. We can substitute these values into the equation 𝑠 equals 𝑢𝑡 plus a half 𝑎𝑡 squared. This gives us 57 is equal to 𝑣 multiplied by four plus a half multiplied by 𝑎 multiplied by four squared. Simplifying this gives us 57 is equal to four 𝑣 plus eight 𝑎.

We now have two simultaneous equations: one for the time period 𝑡 equals zero to 𝑡 equals four and one for the time period between 𝑡 equals four and 𝑡 equals eight. We can use these equations to calculate the acceleration and the velocity at 𝑡 equals four. Adding equation one and equation two gives us eight 𝑣 is equal to 112. Dividing both sides of this equation by eight gives us a value of 𝑣 of 14 meters per second. The velocity of the body at 𝑡 equals four is 14 meters per second.

Subtracting equation one from equation two gives us 16𝑎 is equal to two. Dividing both sides of this equation by 16 gives us a value of 𝑎 of one-eighth meter per second squared. The uniform acceleration of the body is one-eighth meter per second squared.

Our next step is to calculate the initial velocity of the body. In order to do this, we’ll consider the time between 𝑡 equals zero and 𝑡 equals four. Our displacement is 55 meters. Our initial velocity is 𝑢. Our final velocity is 14 meters per second. Our acceleration is one-eighth meter per second squared. And our time is four seconds. Using the equation 𝑣 equals 𝑢 plus 𝑎𝑡 gives us 14 is equal to 𝑢 plus an eighth multiplied by four. One-eighth multiplied by four is 0.5. Therefore, 14 equals 𝑢 plus 0.5. Subtracting 0.5 from both sides of this equation gives us an initial velocity 𝑢 of 13.5 meters per second.

We now have enough information to consider the first 10 seconds of the motion. The initial velocity 𝑢 is 13.5. Our acceleration is one-eighth. 𝑡 equals 10 and 𝑠 — the total displacement — is the unknown. Using the equation 𝑠 equals 𝑢𝑡 plus a half 𝑎𝑡 squared gives us 𝑠 is equal to 13.5 multiplied by 10 plus a half multiplied by an eighth multiplied by 10 squared. This gives us a value of 𝑠 of 141.25.

Therefore, the total distance covered by the body in the first 10 seconds of its motion is 141.25 meters.