Question

a bus is moving at 54 km/ h . drive applied brake and speed reduce to 9 km / h in 5 sec .find distance traveled during this time​

Answer 1

Answer:

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Explanation:

First, convert 54km/h to m/s. This equals 15m/s.

Secondly, I assume that the bus breaks with constant acceleration. When this is true, theres a kinematic equation that says:

v=v0⋅a⋅tv=v0⋅a⋅t  

Which can be rewritten as:

a=v−v0ta=v−v0t  

And we know the following values:

v=0v=0  

v0=15m/sv0=15m/s  

t=8st=8s  

So we get an acceleration of:

a=−15m/s8s=−1.875m/s2a=−15m/s8s=−1.875m/s2  

When we have the acceleration, we can calculate the distance with the kinematic equation (for constant acceleration):

xf=x0+v0⋅t+12a⋅t2xf=x0+v0⋅t+12a⋅t2  

Where we calculate the total distance. We know every value on the right hand side, so we get:

xf=0+15m/s⋅8s+12(−1.875m/s2)⋅(8s)2=60mxf=0+15m/s⋅8s+12(−1.875m/s2)⋅(8s)2=60m  

So it takes the bus 60 meters to break at a constant acceleration as calculated above.