The car, moving uniformly accelerated, through t = 10 s after the start of movement, reached a speed of V = 36 km / h

The car, moving uniformly accelerated, through t = 10 s after the start of movement, reached a speed of V = 36 km / h. Determine the acceleration with which the car was moving. What way did it go? Which way did the car go in the last second?

Find the acceleration of the car by expressing it from the formula:

vx = v0 + atx;

The initial speed of the car v0 according to the conditions of the problem is equal to 0, therefore:

a = v / tx = 36/10 = 3.6 m / s2;

The traversed path will be:

S = v0t + atx ^ 2/2 = 0 + 3.6 * 10 ^ 2/2 = 180 m;

To find the distance traveled in the last second, we will find the speed that the car will pick up in 9 seconds.

v = 3.6 * 9 = 32.4 m / s;

S ‘= 32.4 * 1 + 3.6 * 12/2 = 34.2 m.

Answer: a = 3.6 m / s2; S = 180 m; S ‘= 34.2 m.