# A car moving along a straight highway with a speed of 126 km/hr is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform) and how long does it take for the car to stop?

By BYJU'S Exam Prep

Updated on: September 25th, 2023

The retardation of the car is – 3.06 m/s2 and the time taken by the car to stop is t = 11.43 s. Steps to Calculate the distance traveled by car to stop:

Step 1: Given that

Distance (s) = 200 m

Initial velocity (u) = 126 km/hr

= 126/3600 x 1000

= 35 m/s

Final velocity (v) Equals 0 since the car will eventually come to a stop.

Step 2: Formula used:

3rd Equation of Motion can be written as

v2 = u2 + 2as

Where u is initial velocity, a is acceleration, v is the final velocity and s is the distance

The first Equation of Motion can be written as

v = u + at

Where t is the time taken

Step 3: Calculating retardation of the car using the formula

v2 = u2 + 2as

a = -u2/2s

= – 3.06 m/s2

The car is moving more slowly as indicated by the negative sign. Negative acceleration is what retardation is.

Step 4: Now we have to calculate the time taken,

v = u + at

On rearranging we get

t = -u/a

On substituting we get

t = -35/-3.06

t = 11.43 s

Therefore, the retardation of the car is – 3.06 m/s2, and the time taken by the car to stop is t = 11.43 s

Summary:

## A car moving along a straight highway with a speed of 126 km/hr is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform) and how long does it take for the car to stop?

Within 200 m, an automobile traveling at 126 km/hr on a straight roadway comes to a complete halt. The retardation of the car ‘a’ is – 3.06 m/s2 and the time taken by the car to stop is t is 11.43 s.

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