Question

a bus is moving with a velocity of 90 km/h .its driver applies brakes due to which the bus retards uniformly at the rate of -2m/s^2 and stops after covering certain distance .find the distance covered by the bus during retardation.

Answer 1

Hi there !!
Here's your answer


Given,

Initial velocity (u) = 90km/h = 25m/s [ since the bus is moving]
Retardation (a) = -2m/s²
Final velocity (v) = 0m/s [ since the bus stops after covering a distance]


To find the distance (s) covered by the bus, we'll use the 3rd equation of motion

2as = v² - u²

2(-2)(s) = (0)² - (25)²

-4s = 0 - 625

-4s = -625

$s = \frac{625}{4}$
s = 156.25


Thus,
the distance covered is 156.25m.

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Hope this helps :D

Answer 2

Here, u = 90 km /h =
= 90×5/18 m/s
= 25 m/s
v = 0 m/s
a = -2 m/s^2
distance = s = ?
we know that,
v^2 - u ^2 = 2as.
0^2 - 25^2 = 2× (-2) × S
- 625 = -4 S
S = -625/-4
S= 156.25 km
hence , distance covered is 156 .25 kms

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