A truck is moving at a speed of 72 kmh -1 on a straight road. The driver can produce deceleration of 2 ms -2 by applying brakes. The stopping distance of truck is 13 x m, if the reaction time of the driver is 0.2 s. The value of x is.

Question

A truck is moving at a speed of 72 kmh -1 on a straight road. The driver can produce deceleration of 2 ms -2 by applying brakes. The stopping distance of truck is 13 x m, if the reaction time of the driver is 0.2 s. The value of x is. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

∵ v2=u2+2 a s ⇒ 02=(72 × (5/18))2-2 × 2 s0 ∴ s0=(400/4)=100 m Thus, stopping distance =(72 × (5/18)) × 0.2+s0 =4+100=104=13 x ∴ x=(104/13)=8

Q. A truck is moving at a speed of $72 km h^{- 1}$ on a straight road. The driver can produce deceleration of $2 m s^{- 2}$ by applying brakes. The stopping distance of truck is $13 x m$ , if the reaction time of the driver is $0.2 s$ . The value of $x$ is_____.

$∵ v^{2} = u^{2} + 2 a s$
$\Rightarrow 0^{2} = (72 \times \frac{5}{18})^{2} - 2 \times 2 s_{0}$
$∴ s_{0} = \frac{400}{4} = 100 m$
Thus, stopping distance
$= (72 \times \frac{5}{18}) \times 0.2 + s_{0}$
$= 4 + 100 = 104 = 13 x$
$∴ x = \frac{104}{13} = 8$

Q. A truck is moving at a speed of 72kmh−1 on a straight road. The driver can produce deceleration of 2ms−2 by applying brakes. The stopping distance of truck is 13xm, if the reaction time of the driver is 0.2s. The value of x is_____.

∵v2=u2+2as ⇒02=(72×185​)2−2×2s0​ ∴s0​=4400​=100m Thus, stopping distance =(72×185​)×0.2+s0​ =4+100=104=13x ∴x=13104​=8

Q. A truck is moving at a speed of $72 km h^{- 1}$ on a straight road. The driver can produce deceleration of $2 m s^{- 2}$ by applying brakes. The stopping distance of truck is $13 x m$ , if the reaction time of the driver is $0.2 s$ . The value of $x$ is_____.

$∵ v^{2} = u^{2} + 2 a s$
$\Rightarrow 0^{2} = (72 \times \frac{5}{18})^{2} - 2 \times 2 s_{0}$
$∴ s_{0} = \frac{400}{4} = 100 m$
Thus, stopping distance
$= (72 \times \frac{5}{18}) \times 0.2 + s_{0}$
$= 4 + 100 = 104 = 13 x$
$∴ x = \frac{104}{13} = 8$