Question

when brakes are applied to a bus the retardarion produced is 25 cm s^-2 and the bus takes 20 sec to stop. Calculate the initial velocity of bus. The distance travelled by bus during this time.

Answer 1

$\textbf{Answer:}$

We know that,

$\textbf{v = u - at}$

0 = u - 0.25 × 20

-u = -5

u = 5m/s

Now,

s = ut - 1/2at²

s = 5 × 20 - 1/2 × 0.25 × 20 × 20

s = 100 - 50

s = 50 metres

Therefore, the distance travelled by the bus during this time is $\textbf{50 metres.}$
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Answer 2

HELLO THERE.................
A(ACCELERATION) = 25 CM S ^-2 = 25/100 M = 0.25 M /S^-2
T(TIME) = 20 SEC
U(INITIAL VELOCITY) = ?
V(FINAL VELOCITY) = 0 
FROM THE FIRST EQUATION OF MOTION
V = U + AT 
0 = U + 0.25 ×20
0 = U + 5.00
U = 5 M /S
NOW DISTANCE (S) = ?
USING THE THIRD EQUATION OF MOTION
2AS = V^2 - U ^2
2×0.25×S  = 0 SQUARE - 5 SQUARE

0.5 S = 25
S = 25 ÷0.5 = 25×10/5 = 50 M 
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HOPE IT HELPS :)