Answer to Question #149164 in Mechanics | Relativity for Maryam

Question

Cars moves under constant acceleration covers the distance between 2 points 50m .apart in 5s.its speed as it passes the second point is 15ms^-1(a)velocity at 1st point (b) acceleration

Answer 1

Car moves 50m distance betwen two points in 5s. It gets 15$\frac ms$ speed at scond point (at the end of the way). The car has some acceleration and In whole way acceleration is constant. So we can say it is straight line accelerating motion. We have these formulas:

a = $\frac {v-vo}{t}$ ; (1)

S = v_o t + $\frac a 2$  $\times$ t²; (2)

we have: S = 50 m, v = 15 $\frac ms$, t = 5s.

We change a which in (2) formula with (1). Then we have:

S = v_o $\times$ t + $\frac {v-vo}{2\times t}$ $\times$ t²  = S = v_o t + $\frac {v-vo}{2}$ $\times$ t; (3)

Firstly we find v_o by (3) formula:

50 = v_o $\times$ 5 + $\frac{15-vo}{2}$  $\times$ 5

100 = 10 $\times$ v_o - 5 $\times$ v_o + 75

v_o = 5 ($\frac{m}{s}$ ).

Then we use (1) formula:

a = $\frac{15-5}{2}$ = 5 ($\frac{m}{s^2}$ )

So the acceleration of the car a is 5 $\frac{m}{s^2}$