Question

The position of a particle moving on X axis is given by x= At^2 + Bt + C .The numerical value of A B and C are 7 (-2) and 5 respectively and SI units are used. Find
1) the average velocity during the interval t = 0 to t = 5
2) the average acceleration during the interval t = 0 to t = 5

Answer 1

Answer:

The average velocity is 33 m/s and the average acceleration is 14 m/s².

Explanation:

Given that,

A = 7

B= -2

C = 5

The position of a particle on X axis is

$x=At^2+Bt+C$....(I)

The position at t = 0

$x_{1}=5\ m$

The position at t = 5

Put the value of A, B and C in equation (I)

$x_{2} = 7\times(5)^2+(-2)\times5+5$

$x_{2} =170\ m$

(I). The average velocity during the interval t = 0 to t= 5.

Formula of average velocity is defined as:

$v_{av}=\dfrac{x_{2}-x_{1}}{t_{2}-t_{1}}$

$v_{av}=\dfrac{170-5}{5-0}$

$v_{av}=33\ m/s$

(II). The average acceleration during the interval t = 0 to t = 5

The average acceleration is

$a = \dfrac{v_{f}-v_{i}}{t_{2}-t_{1}}$

The velocity is

$v =\dfrac{dx}{dt}=2At+B$

The initial velocity at t = 0

$v_{i}=2At+B$

put the value of A and B

$v_{i}= 2\times7\times0+(-2)$

$v_{i}=-2\ m/s$

The final velocity at t = 5

$v_{f}= 2\times7\times5+(-2)$

$v_{f}=68\ m/s$

The average acceleration is

$a = \dfrac{68-(-2)}{5-0}$

$a = 14\ m/s^2$

Hence, The average velocity is 33 m/s and the average acceleration is 14 m/s².

Answer 2

Solution

x=7t2−2t+5

(a) v=dxdt=14t−2

at t=5,v=14×5−2=68m/s

(b) a=dvdt=14m/s2

(c ) Average velocity = displacementTime=x5−x05−0

x5=7(5)2−2(5)+5=170m

x0=7(0)2−2(0)+5=5m

vavg=170−55=33m/s

(d) Average acceleration

=Change in velocityTime interval=v5−v05−0

v5=14×5−2=68m/s

v0=14×0−2=−2m/s

aavg=68−(−2)5−0=14m/s2