Question

A driver is traveling at a speed of 72 km/h in car A when he looks down to text a friend that he is running late. Just before he looks down, he is 80 m from an intersection, and a bicyclist, B, traveling at a constant speed of 8 m/s is 32 m from that same intersection. The light turns red during the 3 s text, and when the driver looks up, he hits the brakes, causing a constant deceleration of 5 m/s2. Determine (a) the distance between the car and the bike when the bike reaches the center of the intersection, (b) the velocity that the car appears to have to the cyclist at this time, (c) where the car stops.

Answer 1

Velocity of the car,
`$v_c$` = 72 km/h

= 20 m/s

Distance from the intersection = 20 m

Velocity of the bicycle,
`$v_b$` = 8 m/s

Distance from the intersection = 32 m

a). The time for the bicycle to reach at the center

`$=\frac{\text{distance}}{\text{velocity}}$`

`$=(32)/(8)=4 \ s$`

The distance travelled by the car during text reading (3 s) =
`$v_c$` x 3

= 20 x 3

= 60 m

Distance travelled by the car after braking and till the bike reaches at the center -- remaining time = 4 - 3

= 1 second

∴
`$s= ut - (1)/(2)at^2$`

`$s= 20 * 1 - (1)/(2)* 0.5 * (1)^2$`

= 17.5 m

So the total distance travelled by the car in 4 s = 60 + 17.5

= 77.5 m

So the distance between the car and the bike when the bike reaches at the intersection = 80 - 77.5

= 2.5 m

b). Speed of the car when the bike is at the intersection :

v = u - at

= 20 - (5 x 1)

= 15 m/s

`$v_c$` = 15
`$\hat j$` m/s

`$v_b$` = 8 cos 10
`$\hat i$` - 8 sin 10
`$\hat j$`

Velocity of the car w.r.t cyclist,

`$\vec{v}_(cb) = \vec v_c - \vec v_b$`

`$= 15 \hat j - (8 \cos 10 \ \hat i - 8 \sin 10 \hat j)$`

`$= -7.87\ \hat i+ 16.3\ \hat j $`

c). When the car stops, the distance travelled by the car is

`$v^2=u^2-2as$`

or
`$s=(u^2)/(2a)$`

`$=(20^2)/(2 * 5)$`

= 40 m

The car applied brake when it was 20 m before the intersection.

So the car will stop 20 m after the intersection.