Question

A particle moves along a straight line with a velocity in millimeters per second given by v = 436 - 15tº wheret is in seconds. Calculate the net displacement As and total distance D traveled during the first 6.7 seconds of motion.

Answer 1

Answer:2584.52 mm

Step-by-step explanation:

Given

Velocity is given=436-15t

V is in mm/s

and we know
`\frac{\mathrm{d} x}{\mathrm{d} t}=v`

`\frac{\mathrm{d} x}{\mathrm{d} t}=436-15t`

`dx=\left ( 436-15t\right )dt`

Integrating both sides we get

`\int_(0)^(x)dx=\int_(0)^(6.7)\left ( 436-15t\right )dt`

`x=\left ( 436t-15\cdot (t^2)/(2)\right )_0^(6.7)`

x=2584.52 mm

Here distance is equal to displacement because from graph area under the v -t graph gives displacement when we consider sign and distance if we take absolute values