Brakes are applied to a truck to produce an acceleration of $- 10 m / s^{2}$ . If the truck takes $5 s$ to stop after applying the brakes, the distance covered by the truck before coming to rest is $250 m$ .

200 m

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125 m

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Solution

The correct option is B
125 m

Here, $v = 0$ , $a = - 10 m / s^{2}$ , $t = 5 s$
Using $v = u + a t$ ,
$0 = u + (- 10 \times 5)$
$\Rightarrow u = 50 m / s$
Also, $v^{2}$ = $u^{2} + 2 a s$
$\Rightarrow 0 = 50^{2} + 2 \times (- 10) \times s$
$\Rightarrow s = \frac{2500}{20} m$
$∴ s = 125 m$
Therefore, the truck stops after $125 m$ .

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Brakes are applied to a truck to produce an acceleration of $- 10 m / s^{2}$ . If the truck takes $5 s$ to stop after applying the brakes, the distance covered by the truck before coming to rest is $250 m$ .

Brakes are applied to a truck to produce an acceleration of –10m/s2. If the truck takes 5s to stop after applying the brakes, the distance covered by the truck before coming to rest is 250 m.

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Brakes are applied to a truck to produce an acceleration of –10m/s2. If the truck takes 5s to stop after applying the brakes, the distance covered by the truck before coming to rest is 250 m.

The correct option is B 125 m Here, v=0 , a=−10m/s2 ​ , t=5s Using v=u+at, 0=u+(−10×5) ⇒u=50 m/s Also, v2 = u2+2as ⇒0=502+2×(−10)×s ⇒s=250020 m ∴s=125 m Therefore, the truck stops after 125 m.

Brakes are applied to a truck to produce an acceleration of $- 10 m / s^{2}$ . If the truck takes $5 s$ to stop after applying the brakes, the distance covered by the truck before coming to rest is $250 m$ .