Question

A truck starts from rest and accelerates at 0.7m/s2. 5 s later, a car accelerates from rest at the same starting point with an acceleration of 2.2m/s2.

a) Where and when does the car catch the truck?
_____ m from the starting point
at _____ seconds from the moment the truck started to accelerate.

b) What are their velocities when they meet?
The truck :
The car :

Answer 1

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Question #55091, Physics / Mechanics | Kinematics | Dynamics |

A truck starts from rest and accelerates at $0.7\,\mathrm{m/s^2}$. 5 s later, a car accelerates from rest at the same starting point with an acceleration of $2.2\,\mathrm{m/s^2}$.

a) Where and when does the car catch the truck?

56.47 m from the starting point

at 12.7 seconds from the moment the truck started to accelerate.

b) What are their velocities when they meet?

The truck: 8.89 m/s

The car: 16.94 m/s

Answer:

a) Let's make an assumption that they meet at the distance $d$ from the starting point in $t$ seconds:

Then, $d = 0.5a_1t^2$, where $a_1$ – the acceleration of the truck, (for the truck) and

$d = 0.5a_2(t - \Delta t)^2 - 0.5a_1\Delta t^2$, where $a_2$ – the acceleration of the car and $\Delta t$ – the difference in starting time (for the car).

Thus,

$$0.5a_1t^2 = 0.5a_2(t - \Delta t)^2 - 0.5a_1\Delta t^2,$$ $$0.5 \times 0.7t^2 = 0.5 \times 2.2(t - 5)^2 - 0.5 \times 0.7 \times 25$$ $$0.35t^2 = 1.1(t - 5)^2 - 8.75$$ $$0.35t^2 - 1.1(t^2 - 10t + 25) = -8.75$$ $$-0.75t^2 + 11t - 27.5 + 8.75 = 0$$ $$t^2 - 14.67t + 25 = 0$$ $$D = 215.21 - 100 = 115.21$$ $$t = [14.67 + 10.73]/2 = 12.7\,\mathrm{s}$$

The distance is: $d = 0.5a_1t^2 = 0.5 \times 0.7 \times 12.7^2\,\mathrm{m} = 56.47\,\mathrm{m}$.

b) The velocities are calculated by the equations:

For the truck: $v = a_1t = 0.7 \times 12.7\,\mathrm{m/s} = 8.89\,\mathrm{m/s}$

For the car: $v = a_2(t - \Delta t) = 2.2 \times (12.7 - 5)\,\mathrm{m/s} = 16.94\,\mathrm{m/s}$