On a long straight road, cars move with uniform velocity V1 everywhere except at the bridge, where they travel at another constant rate of V2 . The picture shows a graph. of the distance l.between the two successive cars against time t. Find the length (in m ) of the bridge.

Question

On a long straight road, cars move with uniform velocity V1 everywhere except at the bridge, where they travel at another constant rate of V2 . The picture shows a graph. of the distance l.between the two successive cars against time t. Find the length (in m ) of the bridge. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

2 text nd car reaches the bridge after 20 s. ∴ v1=(400/20)=20 m / s In this time first car goes 200 m, so v2=(200/20)=10 m / s Length of bridge =50 × 10=500 m

Q. On a long straight road, cars move with uniform velocity $V_{1}$ everywhere except at the bridge, where they travel at another constant rate of $V_{2} .$ The picture shows a graph. of the distance $l$ .between the two successive cars against time $t$ . Find the length (in $m$ ) of the bridge.

$2^{nd}$ car reaches the bridge after $20 s$ .
$∴ v_{1} = \frac{400}{20} = 20 m / s$
In this time first car goes $200 m$ , so
$v_{2} = \frac{200}{20} = 10 m / s$
Length of bridge $= 50 \times 10 = 500 m$

Q. On a long straight road, cars move with uniform velocity V1​ everywhere except at the bridge, where they travel at another constant rate of V2​. The picture shows a graph. of the distance l.between the two successive cars against time t. Find the length (in m ) of the bridge.

2nd car reaches the bridge after 20s. ∴v1​=20400​=20m/s In this time first car goes 200m, so v2​=20200​=10m/s Length of bridge =50×10=500m

Q. On a long straight road, cars move with uniform velocity $V_{1}$ everywhere except at the bridge, where they travel at another constant rate of $V_{2} .$ The picture shows a graph. of the distance $l$ .between the two successive cars against time $t$ . Find the length (in $m$ ) of the bridge.

$2^{nd}$ car reaches the bridge after $20 s$ .
$∴ v_{1} = \frac{400}{20} = 20 m / s$
In this time first car goes $200 m$ , so
$v_{2} = \frac{200}{20} = 10 m / s$
Length of bridge $= 50 \times 10 = 500 m$