Question

if x=t³+3t²-2t+2 find distance velocity and acceleration in 1 second​

Answer 1

Given,

Displacement varies with the equation,

x = t³ + 3t² - 2t + 2

Distance at t = 1,

x = 1 + 3*1 - 2*1 + 2

x = 6 - 2 = 4

So, distance at t = 1 is 4 m.

Velocity is rate of change of dispalcement.

v = d(t³ + 3t² - 2t + 2) / dx

v = 3t² + 6t - 2

at, t = 1, velocity = 3*1 + 6*1 - 2 = 3 + 6 - 2 = 7 m/s

Therefore, at t = 1, velocity = 7 m/s

Acceleration is rate of change of velocity.

or, a = dv / dt

or, a = d (3t² + 6t - 2) / t

or, a = 6t + 6

at t = 1, acceleration = 6*1 + 6 = 12 m/s²

Answer 2

Answer:-

Given:-

x = t³ + 3t² - 2t + 2

We have to find;

Distance , Velocity and Acceleration at t = 1 s.

Substitute t = 1 to find the distance.

⟹ x = (1)³ + 3(1)² - 2(1) + 2

⟹ x = 1 + 3 - 2 + 2

⟹ x = 4 m

Now,

Differentiate x with respect to t to find the velocity.

$\implies \sf \: velocity(v) = \dfrac{dx}{dt} \\ \\ \\ \implies \sf \:v = \frac{d( {t}^{3} + 3 {t}^{2} - 2t + 2)}{t} \\ \\ \\ \implies \sf \:v = 3 {t}^{3 - 1} + 2 \times 3 {t}^{2 - 1} - 1 \times 2 {t}^{1 - 1} + 0 \\ \\ \\ \bigg( \because \sf \frac{d(constant)}{dx} = 0 \: \: \& \: \: \frac{d( {x})^{n} }{dx} = n \times {x}^{n - 1} \bigg) \\ \\ \\ \implies \sf \:v = 3 {t}^{2} + 6t - 2 \\ \\ \\ \implies \sf \:v = 3 {(1)}^{2} + 6(1) - 2 \\ \\ \\ \implies \sf \:v = 3 + 6 - 2 \\ \\ \\ \implies \boxed{ \sf \:v = 7 \: m {s}^{ - 1} }$

Now,

Differentiate v = 3t² + 6t - 2 with respect to t to find the acceleration.

$\implies\sf \: acceleration(a) = \frac{d(v)}{dt} \\ \\ \\ \implies\sf \:a = \frac{d( 3{t}^{2} + 6t - 2)}{dt} \\ \\ \\ \implies\sf \:a =2 \times 3 {t}^{2 - 1} + 1 \times 6 {(t)}^{1 - 1} - 0 \\ \\ \\ \implies\sf \:a =6t + 6 \\ \\ \\ \implies\sf \:a =6(1) + 6 \\ \\ \\ \implies \boxed{\sf \:a =12 \: ms ^{ - 2} }$

∴

• Distance travelled is 4 m
• Velocity is 7 m/s.
• Acceleration is 12 m/s².