Question

The velocity of a ball in the horizontal direction is given as the following function:v(t) = 20 + 3t3,where t is in seconds and the velocity is in meters per second.Part (a) Find the change in the position, in meters, between t = 0 and t = 5.5 s.Part (b) Find the acceleration in the horizontal direction, in meters per square second, at t = 0.Part (c) Find the acceleration in the horizontal direction, in meters per square second, at t = 5.5 s.

Answer 1

Answer:

(a) 796.29 m.

(b) 0.

(c) 272.25 m/s^2

Explanation:

The position and acceleration functions can be derived from the velocity function by taking the integral and derivative, respectively.

$x(t) = \int {v(t)} \, dt = \int {(20 + 3t^3)} \, dt = 20t + \frac{3t^4}{4} + C$

$a(t) = \frac{dv(t)}{dt} = 9t^2$

(a) at t = 0: x(0) = C.

at t = 5.5: $x(5.5) = 20(5.5) + \frac{3(5.5)^4}{4} + C = 796.29 + C$

So, the change in the position between t = 0 and t = 5.5 is 796.29 m.

(b) at t = 0: a(0) = 0.

(c) at t = 5.5: $a(5.5) = 9(5.5)^2 = 272.25~m/s^2$

Answer 2

The change in the position of the ball is 796.2968 m while the at time t=5.5 sec, acceleration of the ball is 181.5 m/s².

Given to us

v =  20+3t³

Find the change in the position of the ball?

We know that velocity can be written as,

$v=\dfrac{ds}{dt}$

$ds =v\ dt$

where v is the function of velocity and s in the displacement,

Integrate both sides,

$\int ds= \int v\ dt$

$\int ds= \int (20+3t^3)\ dt$

$s = 20t+3\dfrac{t^4}{4} + c$..... equation of displacement

Substitute t= 0,

s₀ = 0 + c

Substitute t = 5.5 s

s₁ = 796.2968 +c

Change in the position of the ball,

s = s₁ - s₀ = 796.2968 m

What is the acceleration of the ball?

We know that the acceleration can be written as,

$a=\dfrac{dv}{dt}$

substitute,

$a=\dfrac{d(20+3t^3)}{dt}$

$a = 3(2t^2)$

Substitute t=0,

$a = 6(0)^2\\\\a = 0\rm\ m/s^2$

Thus, the acceleration of the ball will be 0 m/s².

Find the acceleration in the horizontal direction, in meters per square second, at t = 5.5 s?

$a = 3(2t^2)$

Substitute the value of t,

$a = 6(5.5)^2\\\\a = 181.5\rm\ m/s^2$

Hence, the acceleration of the ball at time t 5.5 sec is 181.5 m/s².

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