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 $250 m$ .

True

No worries! We‘ve got your back. Try BYJU‘S free classes today!

False

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is B
False

Given,
final velocity, $v = 0$ ,
acceleration, $a = - 10 m s^{- 2}$ ,
and time, $t = 5 s$
Using, $v = u + a t$ ,
$0 = u + (- 10 \times 5)$
$\Rightarrow$ Initial velocity, $u = 50 m s^{- 1}$
Also, $v^{2}$ = $u^{2} + 2 a s$
$\Rightarrow 0 = 50^{2} - 2 \times 10 \times s$
$\Rightarrow s = \frac{50 \times 50}{2 \times 10} = 125 m$
Therefore, the truck will stop after $125 m$ .

Suggest Corrections

Similar questions

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, then find the distance covered by the truck before coming to rest.

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.

i didn't even understand what are we supposed to find in this question.

Join BYJU'S Learning Program

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.

The correct option is B False Given, final velocity, v=0, acceleration, a=−10 m s−2, and time, t=5 s Using, v=u+at, 0=u+(−10×5) ⇒ Initial velocity, u=50 m s−1 Also, v2 = u2+2as ⇒0=502−2×10×s ⇒s=50×502×10=125 m Therefore, the truck will stop 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 $250 m$ .