# Secondary Maths Tuition: Kinematics Question

Solution:

acceleration of car $=\frac{12}{6}=2m/s^2$

$\frac{v}{15-5}=2$

$v=2\times 10=20$

Let $T$ be the time (in seconds) when the car overtakes the truck.

Total distance travelled by car at T seconds = area under graph = $\frac{1}{2}(T-5)(2(T-5))$

($2(T-5)$ is the velocity of car at T seconds, it is obtained in the same way as we calculated v.)

Total distance travelled by truck at T seconds = $\frac{1}{2}(6)(12)+\frac{1}{2}(15-6)(12+16)+16(T-15)$

Equating the two distances will lead to a quadratic equation $T^2-26T+103=0$

Solving that gives $T=21.12s$ or $T=4.876s$ (rejected as car only starts at t=5)

$21.12-5=16.1s$ (3 s.f.)

## Author: mathtuition88

http://mathtuition88.com

This site uses Akismet to reduce spam. Learn how your comment data is processed.