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

The value of A would be 4.

Step-by-step explanation:

Given in the question,

Velocity Vms at time t second, V = A(t-0.05t²)

Integrated from 0 to 15

$\int_{0}^{15}A(t-0.05^{2})dt$

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

$15^{2}A[\frac{1}{2}-\frac{0.05}{3}\times 15]$

225A (0.5 - 0.05 × 5)

225A (0.5 - 0.25)

225A (0.25)

56.25A

To calculate the value of A = $\frac{225}{56.25}=4$

A = 4

The value of A would be 4.

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Answer 2

Step-by-step explanation:

Given in the question,

Velocity Vms at time t second, V = A(t-0.05t²)

Integrated from 0 to 15

\int_{0}^{15}A(t-0.05^{2})dt∫

0

15

A(t−0.05

2

)dt

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

2

15

2

−

3

0.05

(15)

3

]

15^{2}A[\frac{1}{2}-\frac{0.05}{3}\times 15]15

2

A[

2

1

−

3

0.05

×15]

225A (0.5 - 0.05 × 5)

225A (0.5 - 0.25)

225A (0.25)

56.25A

To calculate the value of A = \frac{225}{56.25}=4

56.25

225

=4

A = 4