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A particle P moves along the x-axis. The velocity of P is v ms^(-1) at time t seconds, where v = -2t^2 + 16t - 24 for t ≥ 0.

Question

HLPaper 2

A particle P moves along the x-axis. The velocity of P is v ms^(-1) at time t seconds, where v = -2t^2 + 16t - 24 for t ≥ 0.

Find the times when P is at instantaneous rest.

[2]

Verified

Solution

solving $v = 0$ (M1) $t = 2, t = 6$ (A1)

Find the magnitude of the particle's acceleration at 6 seconds.

[3]

Verified

Solution

use of power rule (M1) $\frac{dv}{dt} = -4t + 16$ (A1) $(t = 6)$ (M1) $\Rightarrow a = -8$ (A1) magnitude $= 8 \text{ m s}^{-2}$ (A1)

Find the greatest speed of P in the interval 0 ≤ t ≤ 6.

[2]

Verified

Solution

using a sketch graph of $v$ (M1) $24 \text{ m s}^{-1}$ (A1)

The particle starts from the origin O. Find an expression for the displacement of P from O at time t seconds.

[4]

Verified

Solution

METHOD ONE $x = \int v \,dt$ (M1) attempt at integration of $v$ (M1) $-\frac{2t^3}{3} + 8t^2 - 24t (+c)$ (A1) attempt to find $c$ (use of $t = 0, x = 0$ ) (M1) $c = 0$ (A1) $(x = -\frac{2t^3}{3} + 8t^2 - 24t)$ (AG)

METHOD TWO $x = \int_0^t v \,dt$ (M1) attempt at integration of $v$ (M1) $[-\frac{2t^3}{3} + 8t^2 - 24t]_0^t$ (A1) attempt to substitute limits into their integral (M1) $x = -\frac{2t^3}{3} + 8t^2 - 24t$ (AG)

Find the total distance travelled by P in the interval 0 ≤ t ≤ 4.

[3]

Verified

Solution

$\int_0^4 |v| \,dt$ (M1)(A1) Note: Award M1 for using the absolute value of $v$ , or separating into two integrals, A1 for the correct expression. $= 32 \text{ m}$ (A1)

A particle P moves along the x-axis. The velocity of P is v ms^(-1) at time t seconds, where v = -2t^2 + 16t - 24 for t ≥ 0.

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