Question

A vehicle is moving in a straight line. The velocity Vms^-1 at time t seconds after the vehicle starts is given by V = A (t− 0.05t^2) for 0 ≤ t ≤ 15. What is the value of A?

Answer 1

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Answer 2

The value of A is 4.

To find : Value of A.

Given :

A vehicle is moving in a straight line.

The velocity $Vms^-^1$ at time "t" seconds.

After the vehicle starts is given by V = A( t - 0.05 t² ).

Limits : 0 ≤ t ≤ 15.

$t_1$ = 0 ; $t_2$ = 15

Integrating the equation " V = A ( t - 0.05 t² ) ", in below formula we get

$\int\limits^{t_2}_{t_1} x.dt$

x = V = A( t - 0.05 t² )

$\int\limits^{15}_0 {A(t - 0.05t^2)} dt$

$\int\limits^{15}_0 {A(\frac{t^{2} }{2} - 0.05\frac{t^3}{3} )}$

${A(\frac{(15)^{2} }{2} - 0.05\frac{(15)^3}{3} )}$

Taking the common term " (15)² ", we get

$(15)^2{A(\frac{1 }{2} - 0.05\times\frac{15}{3} )}$

225 × A ( 0.5 - 0.05 × 5 )

225 × A (0.5 - 0.25)

225 × A (0.25)

56.25 A

To find the value of A:

56.25 A = t²  

A = $\frac{t^2}{56.25}$

A = $\frac{225}{56.25}$

A = 4.

Therefore, the value of A is 4.

To learn more...

1. A vehicle is moving in a straight line. The velocity Vms^-1 at time t seconds after the  vehicle starts is given by V = A(t− 0.05t^2) for 0 ≤ t ≤ 15. What is the value of A?

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2. Find the value of k for which x=0, y=8 is a solution of 3x - 6y = k

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