Answer to Question #267158 in Mechanics | Relativity for Emma's

Question

Velocity of 54m/s acceleration uniformly at rate of 2ms velocity reaches 72m/s

Calculate the distance travelled from the place where the acceleration began

The time taken to cover this distance

Answer 1

Since the distance traveled is $S =\frac {V _ 1 ^ 2-V _ 0 ^ 2} {2\times a}$ , where $V_1$ is the final speed, $V_0$ is the initial, $a$ is the acceleration, then $S =\frac {72 ^ 2-54 ^ 2} {2\times 2} = 567$ m;

Since by another formula the distance is $S = V _ 0\times t +\frac {a\times t ^ 2} {2}$ , we will express from this equation $t$ :

$\frac {a\times t^2}{2}+V_0\times t - S=0$ ,

$\frac {2\times t^2}{2}+54\times t - 567=0$ ,

$t=9$ s.