A big truck moving along a straight line at a speed of $54$ km/hr stop in $5$ s after the breaks are applied. Find the acceleration.

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Solution

$v = u + a t$ (first equation of motion)

Substituting the values

$0 = 15 + a \times 5$

$0 = 15 + 5 a$

$- 15 = 5 a$

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

$- 3 = a$

Therefore acceleration is $- 3 m / s^{2}$

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A big truck moving along a straight line at a speed of 54 km/hr stop in 5s after the breaks are applied. Find the acceleration.

v=u+at (first equation of motion)Substituting the values0=15+a×5 0=15+5a −15=5a −155=a −3=aTherefore acceleration is −3 m/s2

A big truck moving along a straight line at a speed of $54$ km/hr stop in $5$ s after the breaks are applied. Find the acceleration.

$v = u + a t$ (first equation of motion)

Substituting the values

$0 = 15 + a \times 5$

$0 = 15 + 5 a$

$- 15 = 5 a$

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

$- 3 = a$

Therefore acceleration is $- 3 m / s^{2}$