Question

If the speed of a car is reduced from 90 km/hr to 36 km/hr in 5 seconds, what is the distance covered by the car during this time?

Answer 1

Initial speed u = 90 km/hr = 90 * (5/18) = 25 m/s.

Final Speed v = 36 km/hr = 36 * (5/18) = 10 m/s.

Given time t = 5 s.

We know that a = v - u/t

                          = (10 - 25)/5

                          = (-3)

We know that distance traveled S = ut + (1/2) at^2

⇒ 25 * 5 + (1/2) * (-3) * (5)^2

⇒  125 + 25 * (-5.4)

⇒ 125 - 37.5

⇒ 87.5.

Therefore, the distance covered = 87.5 m.

Hope this helps!

Answer 2

$<b><u>Heya mate!! Here's your solution</u>$
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$\huge{Given}$

The speed of a car is reduced from 90 km/h to 36 km/h in 5 seconds.

$\huge{To\: Find}$

The distance covered by the car in the period between the change in speeds.

$\huge{Solution}$

Initial velocity (u) of the car
= 90 km/ h
= 90×1000 m / 60×60 sec
= 90000 m/3600 sec
= 25 m/sec

Final velocity (v) of the car
= 36 km/h
= 36×1000 m / 60×60 sec
= 36000 m/3600 sec
= 10m/s

Time taken = 5 sec
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Now, using the first equation of motion, we find the acceleration of the car.

v = u + at

=> a = (v-u)/t

=> a = (10-25)/5

=> a = (-15)/5

=> a = -3 m/s²

Hence, the acceleration of the car is -3 m/s². The negative sign shows that the car is deacclerating or retarding.
______________________

Now, using the second equation of motion, we find the distance covered by car in this period.

s = ut + ½ at²

=> s = (25×5) + (½ × -3 × 5²)

=> s = (125) + (½ × -3 × 25)

=> s = 125 + (½ × -75)

=> s = 125 + (-37.5)

=> s = 125 - 37.5

=> s = 87.5

Hence, distance covered by car is 87.5 m
_____________________

$\mathcal{\red{Finally,}}$

Your answer is -

$\boxed{\bold{\mathcal{\pink{87.5\: m}}}}$
______________________

$\huge{\bold{\mathfrak{\blue{Thank\: you}}}}$