Question

A mouse runs along a baseboard in your house. The mouse's position as a function of time is given by x(t)=pt 2+qt, with p = 0.36 m/s2 and q = -1.40 m/s .

Answer 1

Answer:

The position of mouse is -1.29 m at 1.5 sec.

Explanation:

Given that,

The value of p = 0.36 m/s²

The value of q = -1.40 m/s²

Suppose, we find the position of mouse at 1.5 sec.

We need to calculate the position

Given equation

$x(t)=pt^2+qt$

Put the value of p and q

$x(t)=0.36t^2+(-1.40)t$

At, t= 1.5 sec

$x(1.5)=0.36\times(1.5)^2-1.40\times1.5$

$x(1.5)=-1.29\ m$

Hence, The position of mouse is -1.29 m at 1.5 sec.

Answer 2

The mouse's position at t = 2 seconds is 1.36 meter.

The mouse's position as a function of time is given by,

                                 $x(t)=pt^{2}+qt$

Substitute p = 0.36 and q = -1.40 in above relation.

                       $x(t)=0.36t^{2}-1.40t$

To find position of mouse at t = 2 second. Substitute t = 2 in above relationship.

                       $x(2)=0.36(2)^{2}-1.40(2)\\\\x(2)=1.44-2.8\\\\x(2)=-1.36 m$

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