Question

The distance travelled by a particle in time t is given
by s = (2.5 m/s ) . Find (a) the average speed of the
particle during the time 0 to 5.0 s and (b) the
instantaneous speed at t=5s​

Answer 1

Answer:

(A) Distance travelled during time 0 to 5 second is,

$s = (2.5) {(5)}^{2}$

$= > 62.5m$

average speed,

$= > \frac{62.5}{5s} = 12.5ms$

(B)

$\frac{ds}{dt} = (2.5)(2t) = 5t$

at t = 5.0s the speed is,

$v = \frac{ds}{dt} = (5)(5) = 25ms$

Answer 2

Explanation:

(A) Distance travelled during time 0 to 5 second is,

s = (2.5) {(5)}^{2}s=(2.5)(5)2

= > 62.5m=>62.5m

average speed,

= > \frac{62.5}{5s} = 12.5ms=>5s62.5=12.5ms

(B)

\frac{ds}{dt} = (2.5)(2t) = 5tdtds=(2.5)(2t)=5t

at t = 5.0s the speed is,

v = \frac{ds}{dt} = (5)(5) = 25msv=dtds=(5)(5)=25ms