Question

a car is moving along a straight highway with speed of 126 km/hr is brought to a stop within a distance of 200 m.what is the retardation of the car and how long does it take for the car to stop

Answer 1

$\sf\: Initial\:velocity\:of\:car = 126 kmph \\ \\ \\ \sf\implies\: 126 × \dfrac {5}{18} \:m/s \\ \\ \\ \large {\boxed {\sf {u = 35 m/s}}} \\ \\ \\ \sf\: Final\:Velocity\: of\: Car = 0 \\ \\ \\ \sf\: Distance\: travelled = 200m \\ \\ \\ \sf\: V^2 = U^2 + 2as \\ \\ \\ \sf\implies\: a = \dfrac {v^2 - u^2}{2s} = - 3.06 m/s^2 \\ \\ \\ \sf\: From\:equation\: of \:motion, v = u + at \\ \\ \\ \sf\: t = \dfrac {v-u}{a} = 11.4 s$

Answer 2

11.4 Seconds

Given:

Speed of the car = 126 km/hr

Converting the speed into m/sec = 126 * 5/18 = 35 meter/sec

s = 200 meters

v = 0

To Find:

t = ?

Solving:

Using the formula v^2 - u^2 = 2as

Substituting the values into the formula we get:

0 - (35)^2 = 2a * 200

- 1225 = 400a

a = - 1225 / 400

a = -3.06 m/sec^2

Now using the formula:

v = u + at

Taking u and a to the other side of the equation we get:

t = v - u / a

Substituting the values in this equation:

= 0 - 35 / - 3.06

= - 35 / - 3.06

= 11.4 seconds

Therefore, it takes 11.4 Seconds for the car to stop.