Question

A particle starts from origin with uniform acceleration its displacement after 2 seconds is given in metres by the relation x=2+5t+7t^2. calculate the magnitude of its ( 1) initial velocity (2) velocity at t=4s (3) uniform acceleration and (4) displacement at t=5s

Answer 1

v = dx/dt

v = d(2+5t+7t²)/dt
v = 5 + 14t

i) initial velocity
= velocity at t = 0
= 5m/s

ii) t = 4
v = 5 + 56
= 61m/s

iii) a = dv/dt
= 14 m/s²

iv) x at t=5
x = 2 + 5×5 + 7×5²
= 2 + 25 + 175
= 202 m


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@di

Answer 2

Answer: The initial velocity is 5 m/s, velocity at 4s is 61 m/s, uniform acceleration is $14\ m/s^2$ and the displacement at 5 s is 202 m.

Explanation:

Given that,

$x=2+5t+7t^2$

We know that,

The velocity is

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

$v = \dfrac{d}{dt}(2+5t+7t^2)$

$v = 5+14t$

The velocity at 4 s

$v = 5+14\times4$

$v = 61 m/s$

The initial velocity is

The velocity at t = 0,

$v = 5 m/s$

The uniform acceleration is

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

$a = \dfrac{d}{dt}(5+14t)$

$a = 14 m/s^2$

The displacement at t = 5s

$x_{5}= 2+25+7\times 25$

$x_{5}= 202 m$

Hence, The initial velocity is 5 m/s, velocity at 4s is 61 m/s, uniform acceleration is $14\ m/s^2$ and the displacement at 5 s is 202 m.