Question

A particle's velocity is described by the function vx = t 2 - 7t + 10 m/s, where t is in s.
a. at what times does the particle reach its turning points?
b. what is the particle's acceleration at each of the turning points?

Answer 1

velocity is given as

$v = t^2 - 7t +10$

when the object reached to its turning its velocity will be zero

$0 = t^2 - 7t +10$

$(t - 5)(t-2) = 0$

so it will reach its turning at two time instant

$t_1 = 2 s and t_2 = 5 s$

Part b)

For acceleration we know that

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

$a = \frac{d}{dt}(t^2 - 7t + 10)$

$a = 2t - 7$

now at t = 2

$a_1 = 2*2 - 7 = -3 m/s^2$

also at t=5

$a_2 = 2*5 - 7 = 3 m/s^2$

Answer 2

The turning point of the particle is at 3.5 seconds with an acceleration of 0.

Acceleration

Acceleration is the time rate of change of velocity. It is given by the equation:

Acceleration = change in velocity / change in time

Given the velocity equation:

v = t² - 7t + 10

The turning point is at dv/dt = 0, hence:

dv/dt = 2t - 7

2t - 7 = 0

t = 3.5 seconds

Acceleration (a) = dv/dt

a = 2t - 7

a(3.5) = 2(3.5) - 7 = 0

The turning point of the particle is at 3.5 seconds with an acceleration of 0.

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