Question

an object moving with constant acceleration along a horizontal path covers the distance between two points 60m apart in 6.0 s. it's speed as it passes the second point is 15 m/s find the speed at the first point and it's acceleration

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

v = 15 m/s. (Final point velocity)

S = 60 m.

t = 6 seconds.

u = ?(Initial point velocity)

$\huge{\bold{\underline{Explanation:-}}}$

From First Kinematic equation.

$\large{\bold{v = u + at}}$

rearranging.

$\large{a = \frac{v - u}{t} ---(1)}$

Now, Using second kinematic equation.

$\large{\bold{S = ut + \frac{1}{2} at^2}}$

Substituting the values of acceleration in the equation , time and distance value.

$\large{60 = u \times 6 + \frac {1}{2} \times \frac{v - u}{ \cancel{6}} \times 6^{ \cancel{2}}}$

$\large{60 = u \times 6 + \frac {1}{ \cancel{2}} \times {v - u} \times \cancel{6}}$

$\large{60 = 6u + (15 - u ) \times 3}$

$\large{60 = 6u + 45 - 3u}$

$\large{60 - 45 = 3u}$

$\large{u = \frac{15}{3}}$

$\large{u = 5 m/s}$

$\huge{\boxed{\boxed{u = 5 m/s}}}$

For acceleration:-

Substituting the values in equation (1).

$\large{a = \frac{15 - 5}{6}}$

$\large{a = \frac{10}{6}}$

$\large{a = 1.67 m/s}$

$\huge{\boxed{\boxed{a = 1.67 m/s^2}}}$

So, the velocity at first point is 5 m/s and acceleration is 1.67 m/s².