Answer:
Velocity will be -10 m/sec
Acceleration will be [tex]-2m/sec^2[/tex]
Step-by-step explanation:
We have given equation of distance [tex]s=10t-t^2[/tex]
Time is given as [tex]0\leq t\leq 10[/tex]
We have to find velocity and acceleration at end of time interval means at t = 10 sec
We know that velocity is rate of change of cdistnce
So velocity [tex]v=\frac{ds}{dt}=\frac{d(10t-t^2)}{dt}=10-2t[/tex]
At t = 0 sec
Velocity [tex]v=10-2\times 10=-10m/sec[/tex]
We know that acceleration is rate of change of velocity
So [tex]a=\frac{dv}{dt}=\frac{d(10-2t)}{dt}=-2m/sec^2[/tex]
