The acceleration of a bus is given by a-at, where a= 1.5 m/s^2...?

Question

The acceleration of a bus is given by a-at, where a= 1.5 m/s^2.

If bus's velocity at time t=1.0 s is 5.0 m/s, what is it's velocity at time t=2.0 s? If the bus's position at time t=1.0 s is 6.0m, what is it's position at time t=2.0 s?

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Answer 1 · 2018-07-15T09:50:01

Please see the explanation below

The acceleration is $a = 1.5 m s^{- 2}$ $a=1.5ms^-2$

The velocity $u_{1} = 5.0 m s^{- 1}$ $u_1=5.0ms^-1$ at $t = 1 s$ $t=1s$

Apply the equation of motion

$v = u + a t$ $v=u+at$

$v = u_{1} + a (t_{2} - t_{1})$ $v=u_1+a(t_2-t_1)$

$= 5 + 1.5 \cdot (2 - 1)$ $=5+1.5*(2-1)$

$= 6.5 m s^{- 1}$ $=6.5ms^-1$

The velocity at $t = 2.0 s$ $t=2.0s$ is $v_{2} = 6.5 m s^{- 1}$ $v_2=6.5ms^-1$

The position is $s_{1} = 6.0 m$ $s_1=6.0m$

Apply the equation of motion

$s_{2} - s_{1} = \frac{1}{2} (v_{2} + v_{1}) (t_{2} - t_{1})$ $s_2-s_1=1/2(v_2+v_1)(t_2-t_1)$

Therefore,

$s_{2} - 6 = \frac{1}{2} (11.5 \cdot 1)$ $s_2-6=1/2(11.5*1)$

$s_{2} = 6 + 5.75 = 11.75 m$ $s_2=6+5.75=11.75m$

The position at time $t = 2 s$ $t=2s$ is $= 11.75 m$ $=11.75m$