A driver takes 0.3 s to fully apply the brakes of truck. The truck moving with velocity of 54 km/h stops in 20 s with a deceleration of $6 m / s^{2}$ . Find the distance for where the driver must apply the brakes without collision from the first obstacle?

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Solution

initial velocity, $u = 54 k m / h = 54 \times \frac{5}{18} = 15 m / s$

distance =speed *time

$= 15 \times 0.2 = 3 m$

$v^{2} - u^{2} = 2 a s$

$0 - 15^{2} = 2 \times (- 6) \times s$

$s = \frac{15^{2}}{12}$

$= 18.75 m$

thus the total distance travelled by driver, $= 18.75 + 3 = 21.75 m$

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A driver takes 0.3 s to fully apply the brakes of truck. The truck moving with velocity of 54 km/h stops in 20 s with a deceleration of 6m/s2. Find the distance for where the driver must apply the brakes without collision from the first obstacle?

initial velocity, u=54km/h=54×518=15m/sdistance =speed *time=15×0.2=3mv2−u2=2as0−152=2×(−6)×ss=15212=18.75mthus the total distance travelled by driver, =18.75+3=21.75m

A driver takes 0.3 s to fully apply the brakes of truck. The truck moving with velocity of 54 km/h stops in 20 s with a deceleration of $6 m / s^{2}$ . Find the distance for where the driver must apply the brakes without collision from the first obstacle?

initial velocity, $u = 54 k m / h = 54 \times \frac{5}{18} = 15 m / s$

distance =speed *time

$= 15 \times 0.2 = 3 m$

$v^{2} - u^{2} = 2 a s$

$0 - 15^{2} = 2 \times (- 6) \times s$

$s = \frac{15^{2}}{12}$

$= 18.75 m$

thus the total distance travelled by driver, $= 18.75 + 3 = 21.75 m$