Question

a car moving with a velocity of 90 km/h stops at 20 sec . find retardation and distance covered
​

Answer 1

u=90km/hr

u=25m/s

v=0

t=20 sec

Therefore

$a = \frac{v - u}{t} \\ = > a = \frac{0 - 25}{20} \\ = > a = - 25 \div 20 \\ = > a = \frac{ - 5}{4} \: \\ s = ut \: + \frac{g {t}^{2} }{2} \\ = 25 \times 20 + \frac{ \frac{ 5}{4} \times {20}^{2} }{2} \\ = 500 + \frac{ \frac{ 5}{4} \times 400 }{2} \\ = 500 + ( 5 \times 50) \\ = 500 + 250 \\ = 750m$

Therefore the required retardation is -5/4 m/s^2 and distance covered is 750 m

Hope it is helpful

Answer 2

Answer:

Retardation is -1 and distance covered is 312.5

Explanation:

Using 1st Equation of Motion ie.

V = u + at

Here, u = 90 km/s = 25 m/s. ; t = 25s ; V = 0

0 = 25 + a (25)

0 = 25 + 25a

-25 = 25a

a = -25/25

Hence, a = -1 m/s

For distance,

Using 2nd Equation of motion,

S = ut + 1/2 (at)*2

S = 25× 25 + 1/2 (-1 × 25)*2

S = 625 + 1/2 ( -625)

S = 625 - 312.5

Hence, S = 312.5 m