Question

when brakes are applied to a truck moving on a straight road with a speed of 90 km per hour it covers a distance of 100m before coming to rest calculate:
1). a uniform retardation of the truck
2). time taken by the truck to stop ​

Answer 1

Answer:

1) v²=u²+2as

0²=25²+2 x a x 100

-625=200a

therefore, retardation a=-625/200=-3.125 m/s²

2) v=u+at

   o=25-3.125t

3.125t=25

therefore, time t=25/3.125=8 s

please mark it brainliest if you are satisfied with the answers

Answer 2

So,

U = 90km/h

V = 0

S = 100

So, by

$v^{2} - u^{2} = 2as$

$-8100 = 2a100$

$\frac{-81}{2} = a$

hence the uniform retardation will be 40.5m/s$^2$

now, time taken to stop

S = UT - 1/2at$^2$

$100 = 90t - \frac{1}{2}\frac{-81}{2}t$

$100= \frac{360t + 81t^{2}}{4}$

$400 = 360t + 81t^{2}$

so, possible values on soving will be

- 5.9 or 0.9

since time can't be negative so time = 0.9