Question 3.4: A car accelerates uniformly from rest to a speed of 100 km/h......
A car accelerates uniformly from rest to a speed of 100 km/h in 18 s.(a) Find the acceleration of the car. (b) Find the distance that the car travels. (c) If the car brakes to a full stop over a distance of 100 m, then find its uniform deceleration.
(a) In this problem we are given v_{o} = 0, v = 100 km/h, and t = 18 s = 5 × 10^{−3} h and we need to find a. So, we can use v = v_{o} + a t to find the acceleration as follows:
a={\frac{v-v_{\mathrm{o}}}{t}}={\frac{100\,{\mathrm{km/h}}-0}{5\times10^{-3}\,{\mathrm{h}}}}=2\times10^{4}\, km/h^2 ≡ 2 \times 10^4 {\frac{1000\,{\mathrm{m}}}{(60\times60\,{\mathrm{s}})^{2}}}=1.54\,{\mathrm{m/s}}^{2}(b) If the car starts from the origin of the x-axis, i.e. x_{o} = 0, then we are given v_{o} = 0, v = 100 km/h, x_{o} = 0, and t = 5 × 10^{−3} h and we need to find x, which in this case equals the distance traveled by the car. So, we use x − x_{o} = \frac{1}{2} (v_{o} + v) t to find the position x as follows:
x=x_{o}+{\frac{1}{2}}(v_{o}+v)\;t=0+{\frac{1}{2}}(0+100\;{\mathrm{km/h}})\times5\times10^{-3}\;{\mathrm{h}}=0.25~{\mathrm{km}}=250~{\mathrm{m}}(c) We are given v_{o} = 100 km/h, v = 0, and x − x_{o} = 0.1 km and we need to find the deceleration a. We use v² = v²_{o} + 2 a (x − x_{o}) to get:
a={\frac{v^{2}-v_{\mathrm{o}}^{2}}{2\,\left(x-x_{\mathrm{o}}\right)}}={\frac{0-(100\,\mathrm{km/h})^{2}}{2\times0.1\,\mathrm{km}}}=-5\times10^{4}\,\mathrm{km/h}^{2}=-3.86\,\mathrm{m/s}^{2}